Question

Show your work, include units, and report the correct number of significant figures for all calculations. 1. The molal freezing point depression constant (K) for p-dichlorobenzene is 7.10°C/m and its normal freezing point (f.p.) is 53.10°C. When 0.209 g of an unknown solute is dissolved in 4.00 g of p-dichlorobenzene, the f.p. is lowered to 50.20°C. Refer to the Introduction and Data Sheet to solve the following problems. °C. a) The f.p. depression (AT) is b) Calculate the molality (m) of the solution. c) Calculate the number of moles of solute in the solution. d) Calculate the molar mass of the unknown solute.

Answer 1

a) T _f is a depression in freezing point = F.P of solvent - F. P of solution

$\therefore$T _f = 53.10 ⁰ C - 50.20 ⁰ C = 2.90 ⁰ C

The f.p depression is 2.90 ⁰ C.

b)

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

Assume unknown solute is non electrolyte, hence i = 1.

We have,  T _f = i $\times$ K _f$\times$ m

$\therefore$ 2.90 ⁰ C = 1 $\times$ 7.10 ⁰ C /m $\times$ molality

$\therefore$ molality =  2.90 ⁰ C / 1 $\times$ 7.10 ⁰ C /m = 0.4084 = 0.408 mol / Kg

ANSWER : Molaltiy of solution = 0.408 mol / kg

C) We have , m = No. of moles of solute / mass of solvent in kg

Here mass of solvent = mass of p-dichlorobenzene = 4.00 g = 0.00400 kg

$\therefore$ 0.4084 mol / kg = No. of moles of solute / 0.00400 kg

$\therefore$ No. of moles of solute = 0.4084 mol / kg $\times$ 0.00400 kg = 0.001634 mol

d) We have , no. of moles = Mass / molar mass

$\therefore$ Molar mass of solute = Mass / no. of moles of solute = 0.209 g / 0.001634 mol = 127.9 g / mol

ANSWER : Molar mass of solute = 128 g /mol