Question

The freezing point of 53.48 g of a pure solvent is measured to be 41.91 ºC. When 2.94 g of an unknown solute (assume the van 't Hoff factor = 1.0000) is added to the solvent the freezing point is measured to be 39.86 ºC. Answer the following questions ( the freezing point depression constant of the pure solvent is 7.33 ºC·kg solvent/mol solute).

-What is the molality of the solution?___ m

-How many moles of solute are present? ___mol

-What is the molecular weight of the solute? ___g/mol

-The freezing point of 58.69 g of a pure solvent is measured to be 50.02 ºC. When 2.51 g of an unknown solute (assume the van 't Hoff factor = 1.0000) is added to the solvent the freezing point is measured to be 47.47 ºC. If the freezing point depression constant of the pure solvent is 7.90 ºC·kg solvent/mol solute calculate the molar mass of the solute ___ g/mol

Answer 1

PART 1

Given

1) Mass of solute = 2.94 g

2) Molar mass of Solute =?

3) Mass of solvent X = 53.48 g = 0.05348 kg

4 ) F.P of solvent X = 41.91 ⁰ C

5 ) F.P of solution = 39.86 ⁰ C

6) i = 1.0000

We have relationship between depression in F.P and molality of solution as, T _f = i K _f m

Where,i is a van't Hoff factor ,   T _f is a depression in freezing point = F.P of solvent - F. P of solution

K _f is a freezing point constant of solvent & m is a molality of solution.

41.91 ⁰ C - 39.86 ⁰ C =1.0000 7.33 ⁰ C / m molality of solution.

2.05 ⁰ C = 7.33 ⁰ C / m molality of solution.

molality of solution= 2.05 ⁰ C / 7.33 ⁰ C / m

molality of solution= 0.2796 m

ANSWER : molality of solution= 0.280 m

We know that, Molality = No. of moles of solute / mass of solvent in kg

No. of moles of solute = molality of solution mass of solvent in kg

No. of moles of solute = 0.2796 mol / kg 0.05348 kg

No. of moles of solute = 0.01495 mol

ANSWER : moles of solute = 0.0150 mol

We have , No. of moles = Mass / Molar mass

Molar mass = Mass / No. of moles

Molar mass = 2.94 g / 0.01495 mol

Molar mass = 196.6 g / mol

ANSWER : Molar mass of unknown solute = 197 g /mol

PART 2

Given

1) Mass of solute = 2.51 g

2) Molar mass of Solute =?

3) Mass of solvent X = 58.69 g = 0.05869 kg

4 ) F.P of solvent X = 50.02 ⁰ C

5 ) F.P of solution = 47.47 ⁰ C

6) i = 1.0000

We have relationship between depression in F.P and molality of solution as, T _f = i K _f m

Where,i is a van't Hoff factor ,   T _f is a depression in freezing point = F.P of solvent - F. P of solution

K _f is a freezing point constant of solvent & m is a molality of solution.

50.02 ⁰ C - 47.47 ⁰ C =1.0000 7.90 ⁰ C / m molality of solution.

2.55 ⁰ C = 7.90 ⁰ C / m molality of solution.

molality of solution= 2.55 ⁰ C / 7.90 ⁰ C / m

molality of solution= 0.3227 m

ANSWER : molality of solution= 0.323 m

We know that, Molality = No. of moles of solute / mass of solvent in kg

No. of moles of solute = molality of solution mass of solvent in kg

No. of moles of solute = 0.3227 mol / kg 0.05869 kg

No. of moles of solute = 0.01894 mol

ANSWER : moles of solute = 0.0189 mol

We have , No. of moles = Mass / Molar mass

Molar mass = Mass / No. of moles

Molar mass = 2.51 g / 0.01894 mol

Molar mass = 132.5 g / mol

ANSWER : Molar mass of unknown solute = 133 g /mol