Question

Vapour pressure of a pure liquid X is 2 atm at 300K. 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

Answer 1

Answer: (A)

20

Answer 2

$\frac { \mathrm { P } ^ { \mathrm { o } } - \mathrm { P } _ { \mathrm { A } } } { \mathrm { P } _ { \mathrm { A } } ^ { \mathrm { o } } } = \frac { \mathrm { n } _ { \mathrm { B } } } { \mathrm { n } _ { \mathrm { A } } }$

$\Rightarrow \mathrm { M } _ { \mathrm { B } } = \frac { 200 \times 2 } { 20 } = 20$