An aqueous solution of $20 %$ (wt/wt) non-volatile solute exerts a pressure of $1.004$ bar at the boiling point of the solvent. What is the molecular mass of the solute?

Open in App

Solution

By roult's law
$\frac{P_{0} - P_{s}}{P_{0}} = \frac{n_{2}}{n_{1} + n_{2}}$
$= \frac{n_{2}}{n_{1}} = \frac{\frac{W_{2}}{M_{2}}}{\frac{W_{1}}{M_{1}}}$
$\frac{P_{0} - P_{s}}{P_{0}} = \frac{W_{2}}{M_{2}} \times \frac{M_{1}}{W_{1}}$
$P_{s} = 1.004 b a r, P_{0} = 1 a t m = 1.013 b a r, W_{2} = 2 g, W_{1} = 98, M_{1} = 188 g m o l^{- 1}$
Put the values
$\frac{1.013 - 1.004}{1.013} = \frac{2 g}{M_{2}} \times \frac{188 g m o l^{- 1}}{98 g}$
$M_{2} = 41.35 g m o l^{- 1}$

Suggest Corrections

Similar questions

Q. An aqueous solution of

2

% non-volatile solute exerts a pressure of

1.004

bar at the normal boiling point of the solvent. What is the molecular mass of the solute?

QUESTION 2.15

An aqueous solution of $2 %$ non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molar mass of the solute?

An aquase solution of 2% non volatile solute exerts a pressure of 1.004 bar at normal boiling point of solvent. What is the molecular mass of solute (vapour pressure of pure water is 1.013bar

Q. Lowering of vapour pressure of