Question

1.When a solution is made from 32.2 g of an unknown nonelectrolyte dissolved in 151 g of solvent, the solution boils at 83.44 °C. The boiling point of the pure solvent and its Kb are 79.31 °C and 4.47 °C/m, respectively. Calculate the molar mass of the unknown electrolyte in g/mol.

2. Calculate the molar mass (in g/mol) of an unknown nonelectrolyte if 0.898 g dissolved in 268.7 mL of water at 30.13 °C has an osmotic pressure of 68.1 mmHg.

Answer 1

1) Change in boiling point, ∆T = 83.44-79.31

= 4.13 °C

Now, molar mass, M = w 2×Kb×1000/w 1×∆T

Mass of Solute, w2 = 32.2 g

Mass of Solvent, w1 = 151 g

K = 4.47 °C/m

So, M = 32.2×4.47×1000/151×4.13

= 230.8 g/mol

2) Osmotic pressure, p = 68.1 mmHg

= 68.1/760 atm (1atm = 760 mmHg)

= 0.0896 atm

Now, Molar mass, M = w2×R×T×1000/p×V(in ml)

w2 = 0.898 g

R = 0.082 Latm/mol.K

T = 30.13 °C = 30.13+273.15 = 303.28 K

V = 268.7 ml

Molar mass = 0.898×0.082×303.28×1000/0.0896×268.7

= 927.5 g/mol