Calculate the molar mass of a nonvolatile sol | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The relative lowering in vapour pressure is given by Raoult's Law: $\frac{P^\circ - P_s}{P^\circ} = \frac{n_2}{n_1 + n_2} \approx \frac{n_2}{n_1}$

Vedclass · Accepted Answer

$72 	ext{ g mol}^{-1}$

Vedclass · Answer

(C) The relative lowering in vapour pressure is given by Raoult's Law: $\frac{P^\circ - P_s}{P^\circ} = \frac{n_2}{n_1 + n_2} \approx \frac{n_2}{n_1}$ for dilute solutions.Given: $\frac{P^\circ - P_s}{P^\circ} = 0.016$, mass of solute $w_2 = 6.4 	ext{ g}$, mass of solvent $w_1 = 100 	ext{ g}$, molar mass of water $M_1 = 18 	ext{ g mol}^{-1}$.Formula: $\frac{P^\circ - P_s}{P^\circ} = \frac{w_2 	imes M_1}{M_2 	imes w_1}$.Substitute the values: $0.016 = \frac{6.4 	imes 18}{M_2 	imes 100}$.$M_2 = \frac{6.4 	imes 18}{0.016 	imes 100} = \frac{115.2}{1.6} = 72 	ext{ g mol}^{-1}$.

Calculate the molar mass of a nonvolatile solute if $6.4 \text{ g}$ of it dissolved in $100 \text{ g}$ water produces a relative lowering in vapour pressure of $0.016$ at $300 \text{ K}$.

Similar Questions

The vapour pressure of a given liquid decreases if?

At a certain temperature,$5 \ g$ of a non-electrolyte solute is dissolved in $100 \ g$ of water. The vapour pressure of the resulting solution is $2985 \ N/m^2$,and the vapour pressure of pure water is $3000 \ N/m^2$. Calculate the molar mass of the solute.

Derive an equation for a solution that shows the relation between total pressure and the mole fraction of a volatile solute and a volatile solvent,and explain it by plotting a graph.

The vapour pressure will be lowest for

$6 \ g$ of a non-volatile solute $(x)$ is dissolved in $100 \ g$ of water. The relative lowering of vapour pressure of the resultant solution is $0.006$. What is the molar mass (in $g \ mol^{-1}$) of $x$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module