x g of a solute is dissolved in y g of two | vedclass

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

Question

(C) According to Raoult's law,the relative lowering of vapor pressure is given by $\frac{P^o - P}{P^o} = \chi_{solute} = \frac{n_{solute}}{n_{solute}

Vedclass · Accepted Answer

$M_A = 2 \ M_B$

Vedclass · Answer

(C) According to Raoult's law,the relative lowering of vapor pressure is given by $\frac{P^o - P}{P^o} = \chi_{solute} = \frac{n_{solute}}{n_{solute} + n_{solvent}}$.Since $n_{solute} \ll n_{solvent}$,the expression simplifies to $\frac{P^o - P}{P^o} \approx \frac{n_{solute}}{n_{solvent}} = \frac{x/M_{solute}}{y/M_{solvent}} = \frac{x \cdot M_{solvent}}{y \cdot M_{solute}}$.For liquid $A$: $(\frac{P^o - P}{P^o})_A = \frac{x \cdot M_A}{y \cdot M_{solute}}$.For liquid $B$: $(\frac{P^o - P}{P^o})_B = \frac{x \cdot M_B}{y \cdot M_{solute}}$.Given that $(\frac{P^o - P}{P^o})_A = 2 	imes (\frac{P^o - P}{P^o})_B$,we have $\frac{x \cdot M_A}{y \cdot M_{solute}} = 2 	imes \frac{x \cdot M_B}{y \cdot M_{solute}}$.Simplifying this,we get $M_A = 2 \ M_B$.

Similar Questions

The graph between vapour pressure and mole fraction of the solvent for an ideal solution is:

$18 \ g$ of glucose $(C_6H_{12}O_6)$ is added to $178.2 \ g$ of water. The vapour pressure of water for this aqueous solution at $100^\circ C$ is ........ $Torr$.

An aqueous solution is $1.00 \ molal$ in $KI$. Which change will cause the vapour pressure of the solution to increase?

State Raoult's law for a solution containing a non-volatile solute and a volatile solvent and explain it.

The vapor pressure of water at $23\,^{\circ}C$ is $19.8\,mm\,Hg$. What will be the vapor pressure (in $mm\,Hg$) of the resulting solution when $0.1\,mol$ of glucose is dissolved in $178.2\,g$ of water?

Vedclass Test Series

Exam Paper Generator

Online Exam Module