It dissolves in 1000 g of water at 20. Mole fractions of substance and water. Determination of the molar mass of a substance

Example 1. Calculate the osmotic pressure of a solution containing 135 g of glucose C 6 H 12 O 6 in 1.5 liters at 0 0 C.

Solution: Osmotic pressure is determined according to Van't Hoff's law:

See RT

We find the molar concentration of the solution by the formula:

Substituting the value of the molar concentration in the expression of Van't Hoff's law, we calculate the osmotic pressure:

π = C m RT= 0.5 mol / L ∙ 8.314 Pa ∙ m 3 / mol ∙ K ∙ 273 = 1134.86 ∙ 10 3 Pa

Example 2.Determine the boiling point of a solution containing 1.84 g of nitrobenzene C 6 H 5 NO 2 in 10 g of benzene. The boiling point of pure benzene is 80.2 0 С.

Solution: The boiling point of the solution will be by ∆t boiling higher than the boiling point of pure benzene: t bale (solution) = t bale (solvent) + ∆t bale;

According to Raoul's law: ∆t bale = Е ∙ С m ,

where E - ebullioscopic solvent constant (tabular value),

C m- molar concentration of the solution, mol / kg

∆t bale = Е ∙ С m = 1.5 ∙ 2.53 = 3.8 0 C.

t bale (solution) = t bale (solvent) + ∆t bale = 80.2 0 C +3.8 0 C = 84 0 C.

901. A solution containing 57 g of sugar С 12 Н 22 О 11 in 500 g of water boils at 100.72 0 С. Determine the ebullioscopic constant of water.

902. A solution containing 4.6 g of glycerol C 3 H 8 O 3 in 71 g of acetone boils at 56.73 0 С. Determine the ebullioscopic constant of acetone if the boiling point of acetone is 56 0 С.

903. Calculate the boiling point of a solution containing 2 g of naphthalene C 10 H 8 in 20 g of ether, if the boiling point of the ether is 35.6 0 C, and its ebullioscopic constant is 2.16.

904.4 g of the substance is dissolved in 100 g of water. The resulting solution freezes at -0.93 0 С. Determine the molecular weight of the solute.

905. Determine the relative molecular weight of benzoic acid if its 10% solution boils at 37.57 0 С. The boiling point of ether is 35.6 0 С, and its ebullioscopic constant is 2.16.

906. Lowering the freezing point of a solution containing 12.3 g of nitrobenzene C 6 H 5 NO 2 in 500 g of benzene is 1.02 0 С. Determine the cryoscopic constant of benzene.

907. The freezing point of acetic acid is 17 0 С, cryoscopic constant is 3.9. Determine the freezing point of a solution containing 0.1 mol of a solute in 500 g of acetic acid CH 3 COOH.

908. A solution containing 2.175 g of a solute in 56.25 g of water freezes at -1.2 0 С. Determine the relative molecular weight of the solute.

909. At what temperature does a solution containing 90 g of glucose С 6 Н 12 О 6 in 1000 g of water boil?

910. 5 g of substance are dissolved in 200 g of alcohol. The solution boils at 79.2 0 С. Determine the relative molecular weight of the substance if the ebullioscopic constant of alcohol is 1.22. The boiling point of alcohol is 78.3 0 С.

911. An aqueous solution of sugar freezes at -1.1 0 С. Determine the mass fraction (%) of sugar С 12 Н 22 О 11 in the solution.

912. In what mass of water should 46 g of glycerin C 3 H 8 O 3 be dissolved in order to obtain a solution with a boiling point of 100.104 0 C?

913. A solution containing 27 g of a substance in 1 kg of water boils at 100.078 0 С. Determine the relative molecular weight of the dissolved substance.

914. Calculate the mass of water in which 300 g of glycerin C 3 H 8 O 3 should be dissolved to obtain a solution that freezes at - 2 0 C.

915. A solution of glucose in water shows an increase in boiling point of 0.416 ° C. Clean out the drop in the freezing point of this solution.

916. Calculate the freezing point of a 20% solution of glycerin C 3 H 8 O 3 in water.

917. 1.6 g of substance are dissolved in 250 g of water. The solution freezes at -0.2 0 C. Calculate the relative molecular weight of the solute.

918. A solution containing 0.5 g of acetone (CH 3) 2 CO in 100 g of acetic acid gives a decrease in the freezing point by 0.34 0 С. Determine the cryoscopic constant of acetic acid.

919. Calculate the mass fraction (%) of glycerin in an aqueous solution, the boiling point of which is 100.39 0 С.

920. How many grams of ethylene glycol C 2 H 4 (OH) 2 do you need to add for each kilogram of water to prepare antifreeze with a freezing point of -9.3 0 C?

921. A solution containing 565 g of acetone and 11.5 g of glycerin C 3 H 5 (OH) 3 boils at 56.38 0 С. Pure acetone boils 56 0 С. Calculate the ebullioscopic constant of acetone.

922. At what temperature does 4% solution freeze ethyl alcohol C 2 H 5 OH in water?

923. Determine the mass fraction (%) of sugar С 12 Н 22 О 11 in an aqueous solution if the solution boils at 101.04 0 С.

924. Which of the solutions will freeze at a lower temperature: 10% glucose solution С 6 Н 12 О 6 or 10% sugar solution С 12 Н 22 О 11?

925. Calculate the freezing point of 12% aqueous (by weight) glycerin solution C 3 H 8 O 3.

926. Calculate the boiling point of a solution containing 100 g of sucrose C 12 H 22 O 11 in 750 g of water.

927. A solution containing 8.535 g of NaNO 3 in 100 g of water crystallizes at t = -2.8 0 С. Determine the cryoscopic constant of water.

928. To prepare the coolant for 20 liters of water, 6 g of glycerin (= 1.26 g / ml) is taken. What will be the freezing point of the prepared antifreeze?

929. Determine the amount of ethylene glycol C 2 H 4 (OH) 2, which must be added to 1 kg of water to prepare a solution with a crystallization temperature of –15 0 С.

930. Determine the crystallization temperature of a solution containing 54 g of glucose C 6 H 12 O 6 in 250 g of water.

931. A solution containing 80 g of naphthalene C 10 H 8 in 200 g of diethyl ether boils at t = 37.5 0 C, and pure ether at t = 35 0 C. Determine the ebulioscopic constant of the ether.

932. Adding 3.24 g of sulfur to 40 g of benzene C 6 H 6 increased the boiling point by 0.91 0 С. How many atoms are sulfur particles in solution if the ebullioscopic constant of benzene is 2.57 0 С.

933. A solution containing 3.04 g of C 10 H 16 O camphor in 100 g of C 6 H 6 benzene boils at t = 80.714 0 C. (The boiling point of benzene is 80.20 0 C). Determine the ebulioscopic constant of benzene.

934. How many grams of carbamide (urea) CO (NH 2) 2 must be dissolved in 125 g of water for the boiling point to rise by 0.26 0 С. Ebullioscopic constant of water is 0.52 0 С.

935. Calculate the boiling point of a 6% (by weight) aqueous solution of glycerin C 3 H 8 O 3.

936. Calculate the mass fraction of sucrose С 12 Н 22 О 11 in an aqueous solution, the crystallization temperature of which is 0.41 0 С.

937. When dissolving 0.4 g of a certain substance in 10 g of water, the crystallization temperature of the solution dropped by 1.24 0 С. Calculate the molar mass of the dissolved substance.

938. Calculate the freezing point of 5% (by weight) sugar solution C 12 H 22 O 11 in water.

939. How many grams of glucose С 6 Н 12 О 6 should be dissolved in 300 g of water to obtain a solution with a boiling point of 100, 5 0 С?

940. A solution containing 8.5 g of some non-electrolyte in 400 g of water boils at a temperature of 100.78 0 С. Calculate the molar mass of the solute.

941. When 0.4 g of a certain substance is dissolved in 10 g of water, the crystallization temperature of the solution becomes –1.24 0 С. Determine the molar mass of the solute.

942. Calculate the mass fraction of sugar С 12 Н 22 О 11 in the solution, the boiling point of which is 100, 13 0 С.

943. Calculate the crystallization temperature of a 25% (by weight) solution of glycerin C 3 H 8 O 3 in water.

944. The crystallization temperature of benzene is С 6 Н 6 5.5 0 С, the cryoscopic constant is 5.12. Calculate the molar mass of nitrobenzene if a solution containing 6.15 g of nitrobenzene in 400 g of benzene crystallizes at 4.86 ° C.

945. A solution of glycerin C 3 H 8 O 3 in water shows an increase in the boiling point by 0.5 0 С. Calculate the crystallization temperature of this solution.

946. Calculate the mass fraction of urea CO (NH 2) 2 in an aqueous solution, the crystallization temperature of which is –5 0 С.

947. How much water should be dissolved 300 g of benzene С 6 Н 6 to obtain a solution with a crystallization temperature of –20 0 С?

948. Calculate the boiling point of a 15% (by weight) solution of glycerin C 3 H 8 O 3 in acetone, if the boiling point of acetone is 56.1 0 C, and the ebulioscopic constant is 1.73.

949. Calculate the osmotic pressure of the solution at 17 0 C if 1 liter contains 18.4 g of glycerin C 3 H 5 (OH) 3.

950. 1 ml of solution contains 10 15 molecules of solute. Calculate the osmotic pressure of the solution at 0 0 С. What volume contains 1 mol of a dissolved substance?

951. How many molecules of a dissolved substance are contained in 1 ml of a solution, the osmotic pressure of which at 54 0 С is equal to 6065 Pa?

952. Calculate the osmotic pressure of 25% (by weight) sucrose solution C 12 H 22 O 11 at 15 0 C (ρ = 1.105 g / ml).

953. At what temperature will the osmotic pressure of a solution containing 45 g of glucose C 6 H 12 O 6 in 1 liter of water reach 607.8 kPa?

954. Calculate the osmotic pressure of 0.25M sugar solution C 12 H 22 O 11 at 38 0 С.

955. At what temperature will the osmotic pressure of a solution containing 60 g of glucose С 6 Н 12 О 6 in 1 liter reach 3 atm?

956. The osmotic pressure of the solution, the volume of which is 5 liters, at 27 0 С is equal to 1.2 ∙ 10 5 Pa. What is the molar concentration of this solution?

957. How many grams of ethyl alcohol С 2 Н 5 ОН should 1 liter of solution contain so that its osmotic pressure is the same as that of a solution containing 4.5 g of formaldehyde СН 2 О in 1 liter at the same temperature.

958. How many grams of ethyl alcohol С 2 Н 5 ОН should be dissolved in 500 ml of water so that the osmotic pressure of this solution at 20 0 С is equal to 4,052 ∙ 10 5 Pa?

959.200 ml of solution contains 1 g of a solute and at 20 0 С have an osmotic pressure of 0.43 ∙ 10 5 Pa. Determine the molar mass of the solute.

960. Determine the molar mass of a solute if a solution containing 6 g of a substance in 0.5 l at 17 0 С has an osmotic pressure of 4.82 ∙ 10 5 Pa.

961. How many grams of glucose C 6 H 12 O 6 should contain 1 liter of solution for its osmotic pressure to be the same as a solution containing 34.2 g of sugar C 12 H 22 O 11 in 1 liter at the same temperature?

962.400 ml of solution contain 2 g of a solute at 27 0 C. The osmotic pressure of the solution is 1.216 ∙ 10 5 Pa. Determine the molar mass of the solute.

963. A sugar solution C 12 H 22 O 11 at 0 0 C has an osmotic pressure of 7.1 ∙ 10 5 Pa. How many grams of sugar are there in 250 ml of such a solution?

964. 2.45 g of urea are contained in 7 liters of solution. The osmotic pressure of the solution at 0 ° C is 1.317 ∙ 10 5 Pa. Calculate the molar mass of urea.

965. Determine the osmotic pressure of the solution, 1 liter of which contains 3.01 ∙ 10 23 molecules at 0 0 С.

966. Aqueous solutions of phenol C 6 H 5 OH and glucose C 6 H 12 O 6 contain equal masses of solutes in 1 liter. In which of the solutions is the osmotic pressure higher at the same temperature? How many times?

967. A solution containing 3 g of nonelectrolyte in 250 ml of water freezes at a temperature of - 0.348 0 С. Calculate the molar mass of the non-electrolyte.

968. A solution containing 7.4 g of glucose C 6 H 12 O 6 in 1 liter at a temperature of 27 0 C has the same osmotic pressure as a urea solution CO (NH 2) 2. How many g of urea is in 500 ml of solution?

969. The osmotic pressure of a solution, in 1 liter of which contains 4.65 g of aniline C 6 H 5 NH 2, at a temperature of 21 0 C is equal to 122.2 kPa. Calculate the molar mass of aniline.

970. Calculate the osmotic pressure at a temperature of 20 0 C 4% sugar solution C 12 H 22 O 11, the density of which is 1.014 g / ml.

971. Determine the osmotic pressure of a solution containing 90.08 g of glucose С 6 Н 12 О 6 in 4 liters at a temperature of 27 0 С.

972. A solution, with a volume of 4 liters, contains 36.8 g of glycerin (C 3 H 8 O 3) at a temperature of 0 ° C. What is the osmotic pressure of this solution?

973. At 0 0 C, the osmotic pressure of a sucrose solution C 12 H 22 O 11 is 3.55 ∙ 10 5 Pa. What mass of sucrose is contained in 1 liter of solution?

974. Determine the value of the osmotic solution, in 1 liter of which with 0.4 mol of non-electrolyte will be obtained at a temperature of 17 0 C.

975. What is the osmotic pressure of a solution containing 6.2 g of aniline (C 6 H 5 NH 2) in 2.5 liters of solution at a temperature of 21 0 С.

976. At 0 0 C, the osmotic pressure of a sucrose solution C 12 H 22 O 11 is 3.55 ∙ 10 5 Pa. What mass of sucrose is contained in 1 liter of solution?

977. At what temperature will an aqueous solution of ethyl alcohol freeze if the mass fraction of C 2 H 5 OH is equal to 25%?

978. A solution containing 0.162 g of sulfur in 20 g of benzene boils at a temperature of 0.081 0 С higher than pure benzene. Calculate the molecular weight of the sulfur in the solution. How many atoms are there in one sulfur molecule?

979. To 100 ml of a 0.5 mol / L aqueous solution of sucrose С 12 Н 22 О 11 300 ml of water was added. What is the osmotic pressure of the resulting solution at 25 ° C?

980. Determine the boiling and freezing points of a solution containing 1 g of nitrobenzene C 6 H 5 NO 2 in 10 g of benzene. The ebuloscopic and cryoscopic constants of benzene are 2.57 and 5.1 K ∙ kg / mol, respectively. The boiling point of pure benzene is 80.2 0 С, the freezing point is -5.4 0 С.

981. What is the freezing point of a non-electrolyte solution containing 3.01 ∙ 10 23 molecules in one liter of water?

982. Solutions of camphor weighing 0.522 g in 17 g of ether boils at a temperature of 0.461 0 С higher than pure ether. Ebullioscopic ether constant 2.16 K ∙ kg / mol. Determine the molecular weight of camphor.

983. The boiling point of an aqueous solution of sucrose is 101.4 0 С. Calculate the molal concentration and mass fraction of sucrose in the solution. At what temperature does this solution freeze?

984. The molecular weight of the non-electrolyte is 123.11 g / mol. What mass of non-electrolyte should be contained in 1 liter of solution so that the solution at 20 ° C has an osmotic pressure of 4.56 ∙ 10 5 Pa?

985. When dissolving 13.0 non-electrolyte in 400 g of diethyl ether (C 2 H 5) 2 O the boiling point increased by 0.453 K. Determine the molecular weight of the solute.

986. Determine the boiling point of an aqueous solution of glucose if the mass fraction of C 6 H 12 O 6 is equal to 20% (for water Ke = 0.516 K ∙ kg / mol).

987. A solution consisting of 9.2 g of iodine and 100 g methyl alcohol(CH 3 OH), boils at 65.0 0 С. How many atoms are included in the iodine molecule in a dissolved state? The boiling point of alcohol is 64.7 0 С, and its ebullioscopic constant is K e = 0.84.

988. How many grams of sucrose С 12 Н 22 О 11 should be dissolved in 100 g of water in order to: a) lower the crystallization temperature by 1 0 С; b) increase the boiling point by 1 0 С?

989. 2.09 of some substance is dissolved in 60 g of benzene. The solution crystallizes at 4.25 0 С. Set the molecular weight of the substance. Pure benzene crystallizes at 5.5 0 C. The cryoscopic constant of benzene is 5.12 K ∙ kg / mol.

990. At 20 ° C, the osmotic pressure of the solution, 100 ml of which contains 6.33 g of the blood coloring matter - hematin, is equal to 243.4 kPa. Determine the molecular weight of hematin.

991. A solution consisting of 9.2 g of glycerin C 3 H 5 (OH) 3 and 400 g of acetone boils at 56.38 0 С. Pure acetone boils at 56.0 0 С. Calculate the ebullioscopic constant of acetone.

992. The vapor pressure of water at 30 0 С is 4245.2 Pa. What mass of sugar C 12 H 22 O 11 should be dissolved in 800 g of water to obtain a solution, the vapor pressure of which is 33.3 Pa less than the vapor pressure of water? Calculate the mass fraction (%) of sugar in the solution.

993. The vapor pressure of ether at 30 0 С is equal to 8.64 ∙ 10 4 Pa. What amount of non-electrolyte should be dissolved in 50 mol of ether to lower the vapor pressure at a given temperature by 2666 Pa?

994. The decrease in vapor pressure over a solution containing 0.4 mol of aniline in 3.04 kg of carbon disulfide at a certain temperature is equal to 1003.7 Pa. The vapor pressure of carbon disulfide at the same temperature is 1.0133 ∙ 10 5 Pa. Calculate the molecular weight of the carbon disulfide.

995. At a certain temperature, the vapor pressure over a solution containing 62 g of phenol C 6 H 5 O in 60 mol of ether is 0.507 ∙ 10 5 Pa. Find the vapor pressure of the ether at this temperature.

996. The vapor pressure of water at 50 0 С is equal to 12334 Pa. Calculate the vapor pressure of a solution containing 50 g of ethylene glycol C 2 H 4 (OH) 2 in 900 g of water.

997. The pressure of water vapor at 65 0 С is equal to 25003 Pa. Determine the water vapor pressure over a solution containing 34.2 g of sugar C 12 H 22 O 12 in 90 g of water at the same temperature.

998. The vapor pressure of water at 10 0 С is 1227.8 Pa. In what volume of water should 16 g of methyl alcohol be dissolved to obtain a solution, the vapor pressure of which is 1200 Pa at the same temperature? Calculate the mass fraction of alcohol in the solution (%).

999. At what temperature will crystallize an aqueous solution, in which the mass fraction of methyl alcohol is 45%.

1000. A water-alcohol solution containing 15% alcohol crystallizes at - 10.26 0 С. Determine the molar mass of alcohol.

2.10.1. Calculation of the relative and absolute masses of atoms and molecules

The relative masses of atoms and molecules are determined using D.I. Mendeleev's values ​​of atomic masses. At the same time, when carrying out calculations for educational purposes, the values ​​of the atomic masses of elements are usually rounded to whole numbers (with the exception of chlorine, the atomic mass of which is assumed to be 35.5).

Example 1. The relative atomic mass of calcium And r (Ca) = 40; relative atomic mass of platinum А r (Pt) = 195.

The relative mass of a molecule is calculated as the sum of the relative atomic masses of the atoms that make up a given molecule, taking into account the amount of their substance.

Example 2. Relative molar mass of sulfuric acid:

M r (H 2 SO 4) = 2A r (H) + A r (S) + 4A r (O) = 2 · 1 + 32 + 4· 16 = 98.

The values ​​of the absolute masses of atoms and molecules are found by dividing the mass of 1 mole of a substance by the Avogadro number.

Example 3. Determine the mass of one calcium atom.

Solution. The atomic mass of calcium is Ar (Ca) = 40 g / mol. The mass of one calcium atom will be equal to:

m (Ca) = А r (Ca): N A = 40: 6.02 · 10 23 = 6,64· 10 -23 g.

Example 4. Determine the mass of one sulfuric acid molecule.

Solution. The molar mass of sulfuric acid is M r (H 2 SO 4) = 98. The mass of one molecule m (H 2 SO 4) is:

m (H 2 SO 4) = M r (H 2 SO 4): N A = 98: 6.02 · 10 23 = 16,28· 10 -23 g.

2.10.2. Calculation of the amount of substance and calculation of the number of atomic and molecular particles from the known values ​​of mass and volume

The amount of a substance is determined by dividing its mass, expressed in grams, by its atomic (molar) mass. The amount of a substance in a gaseous state under normal conditions is found by dividing its volume by the volume of 1 mol of gas (22.4 liters).

Example 5. Determine the amount of sodium n (Na) in 57.5 g of metallic sodium.

Solution. The relative atomic mass of sodium is Ar (Na) = 23. We find the amount of substance by dividing the mass of metallic sodium by its atomic mass:

n (Na) = 57.5: 23 = 2.5 mol.

Example 6. Determine the amount of nitrogen substance, if its volume at normal conditions. is 5.6 liters.

Solution. The amount of nitrogen substance n (N 2) we find by dividing its volume by the volume of 1 mol of gas (22.4 l):

n (N 2) = 5.6: 22.4 = 0.25 mol.

The number of atoms and molecules in a substance is determined by multiplying the amount of substance of atoms and molecules by Avogadro's number.

Example 7. Determine the number of molecules contained in 1 kg of water.

Solution. We find the amount of water substance by dividing its mass (1000 g) by its molar mass (18 g / mol):

n (H 2 O) = 1000: 18 = 55.5 mol.

The number of molecules in 1000 g of water will be:

N (H 2 O) = 55.5 · 6,02· 10 23 = 3,34· 10 24 .

Example 8. Determine the number of atoms contained in 1 liter (NU) oxygen.

Solution. The amount of oxygen substance, the volume of which under normal conditions is 1 liter is equal to:

n (O 2) = 1: 22.4 = 4.46 · 10 -2 mol.

The number of oxygen molecules in 1 liter (n.u.) will be:

N (O 2) = 4.46 · 10 -2 · 6,02· 10 23 = 2,69· 10 22 .

It should be noted that 26.9 · 10 22 molecules will be contained in 1 liter of any gas under normal conditions. Since the oxygen molecule is diatomic, the number of oxygen atoms in 1 liter will be 2 times more, i.e. 5.38 · 10 22 .

2.10.3. Calculation of the average molar mass of the gas mixture and the volume fraction
the gases it contains

The average molar mass of a gas mixture is calculated on the basis of the molar masses of the gases constituting this mixture and their volume fractions.

Example 9. Assuming that the content (in volume percent) of nitrogen, oxygen and argon in the air, respectively, is 78, 21 and 1, calculate the average molar mass of air.

Solution.

M air = 0.78 · M r (N 2) +0.21 · M r (O 2) +0.01 · M r (Ar) = 0.78 · 28+0,21· 32+0,01· 40 = 21,84+6,72+0,40=28,96

Or about 29 g / mol.

Example 10. Gas mixture contains 12 l of NH 3, 5 l of N 2 and 3 l of H 2 measured at normal conditions. Calculate the volume fraction of gases in this mixture and its average molar mass.

Solution. The total volume of the gas mixture is V = 12 + 5 + 3 = 20 liters. The volume fractions of j gases will be equal:

φ (NH 3) = 12: 20 = 0.6; φ (N 2) = 5: 20 = 0.25; φ (H 2) = 3: 20 = 0.15.

The average molar mass is calculated on the basis of the volume fractions of the gases constituting this mixture and their molecular weights:

M = 0.6 · M (NH 3) +0.25 · M (N 2) +0.15 · M (H 2) = 0.6 · 17+0,25· 28+0,15· 2 = 17,5.

2.10.4. Calculation of the mass fraction of a chemical element in a chemical compound

The mass fraction ω of a chemical element is defined as the ratio of the mass of an atom of a given element X contained in a given mass of a substance to the mass of this substance m. Mass fraction is a dimensionless quantity. It is expressed in fractions of one:

ω (X) = m (X) / m (0<ω< 1);

or percentage

ω (X),% = 100 m (X) / m (0%<ω<100%),

where ω (X) is the mass fraction of a chemical element X; m (X) is the mass of a chemical element X; m is the mass of the substance.

Example 11. Calculate the mass fraction of manganese in manganese oxide (VII).

Solution. The molar masses of the substances are: M (Mn) = 55 g / mol, M (O) = 16 g / mol, M (Mn 2 O 7) = 2M (Mn) + 7M (O) = 222 g / mol. Therefore, the mass of Mn 2 O 7 with the amount of substance 1 mol is:

m (Mn 2 O 7) = M (Mn 2 O 7) · n (Mn 2 O 7) = 222 · 1 = 222 g.

From the formula Mn 2 O 7 it follows that the amount of the substance of the manganese atoms is twice the amount of the substance of the manganese (VII) oxide. Means,

n (Mn) = 2n (Mn 2 O 7) = 2 mol,

m (Mn) = n (Mn) · M (Mn) = 2 · 55 = 110 g.

Thus, the mass fraction of manganese in manganese (VII) oxide is equal to:

ω (X) = m (Mn): m (Mn 2 O 7) = 110: 222 = 0.495 or 49.5%.

2.10.5. Establishing the formula of a chemical compound by its elemental composition

The simplest chemical formula of a substance is determined on the basis of the known values ​​of the mass fractions of the elements that make up this substance.

Suppose there is a sample of a substance Na x P y O z with a mass of m o g. Let us consider how its chemical formula is determined if the quantities of matter of atoms of elements, their masses or mass fractions in a known mass of a substance are known. The formula of a substance is determined by the ratio:

x: y: z = N (Na): N (P): N (O).

This ratio does not change if each of its members is divided by the Avogadro number:

x: y: z = N (Na) / N A: N (P) / N A: N (O) / N A = ν (Na): ν (P): ν (O).

Thus, in order to find the formula of a substance, it is necessary to know the ratio between the amounts of substances of atoms in the same mass of a substance:

x: y: z = m (Na) / M r (Na): m (P) / M r (P): m (O) / M r (O).

If we divide each term of the last equation by the mass of the sample m o, then we get an expression that allows us to determine the composition of the substance:

x: y: z = ω (Na) / M r (Na): ω (P) / M r (P): ω (O) / M r (O).

Example 12. The substance contains 85.71 mass. % carbon and 14.29 wt. % hydrogen. Its molar mass is 28 g / mol. Determine the simplest and true chemical formula of this substance.

Solution. The ratio between the number of atoms in a molecule C x H y is determined by dividing the mass fractions of each element by its atomic mass:

x: y = 85.71 / 12: 14.29 / 1 = 7.14: 14.29 = 1: 2.

Thus, the simplest formula for a substance is CH 2. The simplest formula of a substance does not always coincide with its true formula. In this case, the formula CH 2 does not correspond to the valence of the hydrogen atom. To find the true chemical formula, you need to know the molar mass of a given substance. In this example, the molar mass of the substance is 28 g / mol. Dividing 28 by 14 (the sum of atomic masses corresponding to the formula unit CH 2), we obtain the true ratio between the number of atoms in a molecule:

We get the true formula of the substance: C 2 H 4 - ethylene.

Instead of the molar mass for gaseous substances and vapors, the problem statement can indicate the density for any gas or air.

In the case under consideration, the air density of the gas is 0.9655. Based on this value, the molar mass of the gas can be found:

M = M air · D air = 29 · 0,9655 = 28.

In this expression, M is the molar mass of the gas C x H y, M air is the average molar mass of air, D air is the density of the gas C x H y in air. The resulting molar mass is used to determine the true formula of a substance.

The problem statement may not indicate the mass fraction of one of the elements. It is found by subtracting from one (100%) the mass fractions of all other elements.

Example 13. The organic compound contains 38.71 mass. % carbon, 51.61 wt. % oxygen and 9.68 wt. % hydrogen. Determine the true formula of this substance if its vapor density for oxygen is 1.9375.

Solution. We calculate the ratio between the number of atoms in the molecule C x H y O z:

x: y: z = 38.71 / 12: 9.68 / 1: 51.61 / 16 = 3.226: 9.68: 3.226 = 1: 3: 1.

The molar mass M of a substance is equal to:

M = M (O 2) · D (O 2) = 32 · 1,9375 = 62.

The simplest formula of the substance is CH 3 O. The sum of atomic masses for this formula unit will be 12 + 3 + 16 = 31. We divide 62 by 31 and we get the true ratio between the number of atoms in a molecule:

x: y: z = 2: 6: 2.

Thus, the true formula of the substance is C 2 H 6 O 2. This formula corresponds to the composition of dihydric alcohol - ethylene glycol: CH 2 (OH) -CH 2 (OH).

2.10.6. Determination of the molar mass of a substance

The molar mass of a substance can be determined on the basis of the density of its vapor in a gas with a known value of the molar mass.

Example 14. The vapor density of some organic compound for oxygen is 1.8125. Determine the molar mass of this compound.

Solution. The molar mass of the unknown substance M x is equal to the product of the relative density of this substance D by the molar mass of the substance M, according to which the value of the relative density is determined:

M x = D · M = 1.8125 · 32 = 58,0.

Substances with the found value of the molar mass can be acetone, propionic aldehyde and allyl alcohol.

The molar mass of a gas can be calculated using the standard molar volume.

Example 15. Mass of 5.6 liters of gas at standard. is 5.046 g. Calculate the molar mass of this gas.

Solution. The molar volume of gas at normal conditions is 22.4 liters. Therefore, the molar mass of the target gas is

M = 5.046 · 22,4/5,6 = 20,18.

The sought gas is neon Ne.

The Clapeyron – Mendeleev equation is used to calculate the molar mass of a gas whose volume is given under conditions other than normal.

Example 16. At a temperature of 40 about C and a pressure of 200 kPa, the mass of 3.0 liters of gas is 6.0 g. Determine the molar mass of this gas.

Solution. Substituting the known values ​​into the Clapeyron – Mendeleev equation, we obtain:

M = mRT / PV = 6.0 · 8,31· 313/(200· 3,0)= 26,0.

The gas in question is acetylene C 2 H 2.

Example 17. During the combustion of 5.6 l (NU) of hydrocarbon, 44.0 g of carbon dioxide and 22.5 g of water were obtained. The relative oxygen density of the hydrocarbon is 1.8125. Determine the true chemical formula of the hydrocarbon.

Solution. The reaction equation for the combustion of a hydrocarbon can be represented as follows:

C x H y + 0.5 (2x + 0.5y) O 2 = x CO 2 + 0.5y H 2 O.

The amount of hydrocarbon is 5.6: 22.4 = 0.25 mol. As a result of the reaction, 1 mol of carbon dioxide and 1.25 mol of water are formed, which contains 2.5 mol of hydrogen atoms. When a hydrocarbon is burned with an amount of 1 mol, 4 mol of carbon dioxide and 5 mol of water are obtained. Thus, 1 mol of hydrocarbon contains 4 mol of carbon atoms and 10 mol of hydrogen atoms, i.e. chemical formula of hydrocarbon C 4 H 10. The molar mass of this hydrocarbon is M = 4 · 12 + 10 = 58. Its relative density for oxygen D = 58: 32 = 1.8125 corresponds to the value given in the problem statement, which confirms the correctness of the found chemical formula.

Overload 427.
Calculate the molar fractions of alcohol and water in a 96% (by weight) solution of ethyl alcohol.
Solution:
Mole fraction(N i) - the ratio of the amount of a solute (or solvent) to the sum of the amounts of all
substances in solution. In a system consisting of alcohol and water, the mole fraction of water (N 1) is

And the mole fraction of alcohol , where n 1 is the amount of alcohol; n 2 is the amount of water.

We calculate the mass of alcohol and water contained in 1 liter of solution, provided that their densities are equal to one from the proportions:

a) the mass of alcohol:

b) the mass of water:

We find the amount of substances by the formula:, where m (B) and M (B) are the mass and amount of the substance.

Now let's calculate the mole fractions of substances:

Answer: 0,904; 0,096.

Task 428.
666g KOH is dissolved in 1kg of water; the density of the solution is 1.395 g / ml. Find: a) mass fraction of KOH; b) molarity; c) molality; d) mole fractions of alkali and water.
Solution:
a) Mass fraction- the percentage of the mass of the solute to the total mass of the solution is determined by the formula:

where

m (solution) = m (H2O) + m (KOH) = 1000 + 666 = 1666

b) Molar (volume-molar) concentration shows the number of moles of a solute contained in 1 liter of solution.

Let us find the mass of KOH per 100 ml of solution according to the formula: formula: m = p V, where p is the density of the solution, V is the volume of the solution.

m (KOH) = 1.395 . 1000 = 1395 g.

Now let's calculate the molarity of the solution:

We find how many grams of HNO 3 are there in 1000 g of water, making up the proportion:

d) Mole fraction (N i) - the ratio of the amount of a dissolved substance (or solvent) to the sum of the amounts of all substances in solution. In a system consisting of alcohol and water, the mole fraction of water (N 1) is equal to a mole fraction of alcohol, where n 1 is the amount of alkali; n 2 is the amount of water.

100 g of this solution contains 40 g of KOH 60 g of H2O.

Answer: a) 40%; b) 9.95 mol / l; c) 11.88 mol / kg; d) 0.176; 0.824.

Task 429.
The density of a 15% (by weight) H 2 SO 4 solution is 1.105 g / ml. Calculate: a) normality; b) molarity; c) the molality of the solution.
Solution:
Let's find the mass of the solution by the formula: m = p V where p is the density of the solution, V is the volume of the solution.

m (H 2 SO 4) = 1.105 . 1000 = 1105 g.

We find the mass of H 2 SO 4 contained in 1000 ml of solution from the proportion:

Determine the molar mass of the equivalent of H 2 SO 4 from the ratio:

M E (B) is the molar mass of the acid equivalent, g / mol; M (B) is the molar mass of the acid; Z (B) - equivalent number; Z (acid) is equal to the number of H + ions in H 2 SO 4 → 2.

a) Molar equivalent concentration (or normality) shows the number of equivalents of a solute contained in 1 liter of solution.

b) Molar concentration

Now let's calculate the molality of the solution:

c) Molar concentration (or molality) shows the number of moles of a solute contained in 1000 g of solvent.

We find how many grams of H 2 SO 4 are contained in 1000 g of water, making up the proportion:

Now let's calculate the molality of the solution:

Answer: a) 3.38n; b) 1.69 mol / l; 1.80 mol / kg.

Task 430.
The density of a 9% (by weight) sucrose solution C 12 H 22 O 11 is 1.035 g / ml. Calculate: a) the concentration of sucrose in g / l; b) molarity; c) the molality of the solution.
Solution:
M (C 12 H 22 O 11) = 342 g / mol. Let us find the mass of the solution by the formula: m = p V, where p is the density of the solution, V is the volume of the solution.

m (C 12 H 22 O 11) = 1.035. 1000 = 1035 g.

a) The mass of C 12 H 22 O 11 contained in the solution is calculated by the formula:

where
- mass fraction of the dissolved substance; m (in-va) - the mass of the solute; m (solution) is the mass of the solution.

The concentration of a substance in g / l shows the number of grams (mass units) contained in 1 liter of solution. Therefore, the sucrose concentration is 93.15 g / l.

b) Molar (volume-molar) concentration (CM) shows the number of moles of a solute contained in 1 liter of solution.

v) Molar concentration(or molality) indicates the number of moles of a solute contained in 1000 g of a solvent.

We find how many grams of C 12 H 22 O 11 are contained in 1000 g of water, making up the proportion:

Now let's calculate the molality of the solution:

Answer: a) 93.15 g / l; b) 0.27 mol / l; c) 0.29 mol / kg.

The properties of dilute solutions, which depend only on the amount of non-volatile solute, are called colligative properties... These include lowering the vapor pressure of the solvent over the solution, increasing the boiling point and lowering the freezing point of the solution, and osmotic pressure.

Lowering the freezing point and increasing the boiling point of the solution in comparison with the pure solvent:

T deputy. = = K TO. m 2 ,

T bale. = = K NS. m 2 .

where m 2 - molality of the solution, K To and K E - cryoscopic and ebulioscopic solvent constants, X 2 - mole fraction of solute, H pl. and H isp. - enthalpy of melting and evaporation of the solvent, T pl. and T bale. - the melting and boiling points of the solvent, M 1 - the molar mass of the solvent.

Osmotic pressure in dilute solutions can be calculated using the equation

where X 2 - molar fraction of solute, - molar volume of the solvent. In very dilute solutions, this equation becomes van't Hoff equation:

where C Is the molarity of the solution.

The equations describing the colligative properties of non-electrolytes can also be applied to describe the properties of electrolyte solutions by introducing the Van't Hoff correction factor i, for example:

= iCRT or T deputy. = iK TO. m 2 .

The isotonic coefficient is related to the degree of dissociation of the electrolyte:

i = 1 + (- 1),

where is the number of ions formed during the dissociation of one molecule.

Solubility of a solid in an ideal solution at temperature T described the Schroeder equation:

,

where X- the mole fraction of the solute in the solution, T pl. - melting point and H pl. Is the enthalpy of melting of the solute.

EXAMPLES

Example 8-1. Calculate the solubility of bismuth in cadmium at 150 and 200 o C. The enthalpy of melting of bismuth at the melting temperature (273 o C) is 10.5 kJ. mol –1. Assume that an ideal solution is formed and the enthalpy of melting does not depend on temperature.

Solution. Let's use the formula .

At 150 o C , where X = 0.510

At 200 o C , where X = 0.700

Solubility increases with temperature, which is characteristic of the endothermic process.

Example 8-2. A solution of 20 g of hemoglobin in 1 liter of water has an osmotic pressure of 7.52 10 –3 atm at 25 o C. Determine the molar mass of hemoglobin.

65 kg. mol –1.

TASKS

1. Calculate the minimum osmotic work done by the kidneys to excrete urea at 36.6 o C, if the concentration of urea in the plasma is 0.005 mol. l –1, and in urine 0.333 mol. l –1.
2. 10 g of polystyrene is dissolved in 1 liter of benzene. The height of the solution column (with a density of 0.88 g cm –3) in an osmometer at 25 o C is 11.6 cm. Calculate the molar mass of polystyrene.
3. Human serum albumin has a molar mass of 69 kg. mol –1. Calculate the osmotic pressure of a solution of 2 g of protein in 100 cm 3 of water at 25 o C in Pa and in mm column of the solution. Consider the density of the solution equal to 1.0 g cm –3.
4. At 30 o C, the vapor pressure of the aqueous sucrose solution is 31.207 mm Hg. Art. The vapor pressure of pure water at 30 o C is 31.824 mm Hg. Art. The density of the solution is 0.99564 g cm –3. What is the osmotic pressure of this solution?
5. Human blood plasma freezes at –0.56 o C. What is its osmotic pressure at 37 o C, measured with a membrane that is only permeable to water?
6. * The molar mass of the enzyme was determined by dissolving it in water and measuring the height of the solution column in an osmometer at 20 o C, and then extrapolating the data to zero concentration. The following data were obtained:

C, mg. cm –3
h, cm

7. The molar mass of a lipid is determined by the rise in the boiling point. The lipid can be dissolved in methanol or chloroform. The boiling point of methanol is 64.7 o C, the heat of vaporization is 262.8 cal. g –1. The boiling point of chloroform is 61.5 o C, the heat of vaporization is 59.0 cal. g –1. Calculate the ebulioscopic constants for methanol and chloroform. Which solvent is best to use to determine molar mass with maximum accuracy?
8. Calculate the freezing point of an aqueous solution containing 50.0 g of ethylene glycol in 500 g of water.
9. A solution containing 0.217 g of sulfur and 19.18 g of CS 2 boils at 319.304 K. The boiling point of pure CS 2 is 319.2 K. The ebulioscopic constant of CS 2 is 2.37 K. kg. mol –1. How many sulfur atoms are there in a sulfur molecule dissolved in CS 2?
10. 68.4 g of sucrose is dissolved in 1000 g of water. Calculate: a) vapor pressure, b) osmotic pressure, c) freezing point, d) boiling point of the solution. The vapor pressure of pure water at 20 o C is 2314.9 Pa. Cryoscopic and ebulioscopic constant waters are equal to 1.86 and 0.52 K. kg. mol –1, respectively.
11. A solution containing 0.81 g of hydrocarbon H (CH 2) n H and 190 g of ethyl bromide freezes at 9.47 o C. The freezing point of ethyl bromide is 10.00 o C, cryoscopic constant is 12.5 K. kg. mol –1. Calculate n.
12. When 1.4511 g of dichloroacetic acid is dissolved in 56.87 g of carbon tetrachloride, the boiling point increases by 0.518 degrees. The boiling point of CCl 4 is 76.75 o C, the heat of vaporization is 46.5 cal. g –1. What is the apparent molar mass of the acid? What explains the discrepancy with the true molar mass?
13. A certain amount of a substance dissolved in 100 g of benzene lowers its freezing point by 1.28 o C. The same amount of a substance dissolved in 100 g of water lowers its freezing point by 1.395 o C. The substance has a normal molar mass in benzene, and in water completely dissociated. How many ions does the substance dissociate in an aqueous solution? The cryoscopic constants for benzene and water are 5.12 and 1.86 K. kg. mol –1.
14. Calculate the ideal solubility of anthracene in benzene at 25 o C in terms of molality. The enthalpy of melting of anthracene at the melting point (217 o C) is 28.8 kJ. mol –1.
15. Calculate solubility NS-dibromobenzene in benzene at 20 and 40 o C, assuming that an ideal solution is formed. Enthalpy of melting NS-dibromobenzene at its melting point (86.9 o C) is 13.22 kJ. mol –1.
16. Calculate the solubility of naphthalene in benzene at 25 o C, assuming that an ideal solution is formed. The enthalpy of melting of naphthalene at its melting point (80.0 o C) is 19.29 kJ. mol –1.
17. Calculate the solubility of anthracene in toluene at 25 o C, assuming that an ideal solution is formed. The enthalpy of melting of anthracene at the melting point (217 o C) is 28.8 kJ. mol –1.
18. Calculate the temperature at which pure cadmium is in equilibrium with the Cd - Bi solution, the molar fraction of Cd in which is 0.846. The enthalpy of melting of cadmium at the melting point (321.1 o C) is 6.23 kJ. mol –1.