Vapour pressure of a pure liquid X is 2 atm at 300 K. It is lowered to 1 atm on dissolving 1 g of Y in 20 g of liquid X. If molar mass of X is 200, what is the molar mass of Y ?

Question

Vapour pressure of a pure liquid X is 2 atm at 300 K. It is lowered to 1 atm on dissolving 1 g of Y in 20 g of liquid X. If molar mass of X is 200, what is the molar mass of Y ? (A) 20 (B) 50 (C) 100 (D) 200

Tardigrade Admin · Accepted Answer

(p° - pA/p°A) = (nB/nA) (2-1/2) = (WB/MB)×(MA/WA) ⇒ (1/2) = (1/MB)×(200/20) ⇒ MB = (200 × 2/20 ) = 20

Q. Vapour pressure of a pure liquid $X$ is $2$ atm at $300 K$ . It is lowered to $1$ atm on dissolving $1 g$ of $Y$ in $20 g$ of liquid $X$ . If molar mass of $X$ is $200$ , what is the molar mass of $Y$ ?

$20$

$50$

$100$

$200$

$\frac{p ^{\circ} - p _{A}}{p _{A}^{\circ}} = \frac{n _{B}}{n _{A}}$
$\frac{2 - 1}{2} = \frac{W _{B}}{M _{B}} \times \frac{M _{A}}{W _{A}}$
$\Rightarrow \frac{1}{2} = \frac{1}{M _{B}} \times \frac{200}{20}$
$\Rightarrow M_{B} = \frac{200 \times 2}{20} = 20$

Q. Vapour pressure of a pure liquid X is 2 atm at 300K. It is lowered to 1 atm on dissolving 1g of Y in 20g of liquid X. If molar mass of X is 200, what is the molar mass of Y ?

20

50

100

200