At 300 K, 0.06 kg of an organic solute is d | vedclass

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

Question

(C) For a dilute solution, the relative lowering of vapour pressure is given by: $\frac{P^0 - P_s}{P^0} = \frac{n_2}{n_1} = \frac{W_2 	imes M_1}{M_2

Vedclass · Accepted Answer

$340$

Vedclass · Answer

(C) For a dilute solution, the relative lowering of vapour pressure is given by: $\frac{P^0 - P_s}{P^0} = \frac{n_2}{n_1} = \frac{W_2 	imes M_1}{M_2 	imes W_1}$ Given: $W_2 = 0.06 \ kg = 60 \ g$, $W_1 = 1 \ kg = 1000 \ g$, $P^0 = 3.78 \ kPa$, $P_s = 3.768 \ kPa$, $M_1 = 18 \ g \ mol^{-1}$ (for water). $\Delta P = P^0 - P_s = 3.78 - 3.768 = 0.012 \ kPa$. Rearranging the formula for $M_2$: $M_2 = \frac{W_2 	imes M_1 	imes P^0}{\Delta P 	imes W_1}$ $M_2 = \frac{60 	imes 18 	imes 3.78}{0.012 	imes 1000} = \frac{4082.4}{12} = 340.2 \ g \ mol^{-1}$ The closest integer value is $340 \ g \ mol^{-1}$.

Similar Questions

At $25\,^oC$,the vapour pressure of pure liquid $A$ (mol. wt. $= 40$) is $100\,torr$,while that of pure liquid $B$ is $40\,torr$ (mol. wt. $= 80$). The vapour pressure at $25\,^oC$ of a solution containing $20\,g$ of each $A$ and $B$ is ........... $torr$.

Vapour pressure of $CCl_4$ at $25\,^{\circ}C$ is $143\,mm$ of $Hg$. $0.5\,g$ of a non-volatile solute (mol. wt. $= 65$) is dissolved in $100\,mL$ $CCl_4$. Find the vapour pressure of the solution (Density of $CCl_4 = 1.58\,g/cm^3$).

The vapour pressure of a liquid depends on

Two beakers $A$ and $B$ containing pure water and an aqueous solution,respectively,are placed in a closed container. What will be observed gradually with the passage of time?

Vedclass Test Series

Exam Paper Generator

Online Exam Module