Question

When $2$ g of gas A is introduced into an evacuated flask kept at ${25}^{o}C$, the pressure was found to be $1$ atm. If $3$ g of another gas B is then added to the same flask containing A, the pressure becomes $1.5$ atm. Assuming ideal behaviour, the ratio of molecular weights $(M_A:M_B)$ is :

• A. 1:3
• B. 3:1
• C. 2:3
• D. 3:2

Answer 1

The correct option is A 1:3

$Moles\propto Pressure$
For A, $\frac{2}{M{w}_A}\propto 1$ atm ......(i)

Pressure of $B=1.5-1=0.5$ atm
For B, $\frac{3}{M{w}_B}\propto 0.5$ atm ......(ii)
Dividing (ii) by (i), we get,

$\frac{3}{M{w}_B}\times \frac{M{w}_A}{2}=\frac{0.5}{1}$
$\frac{M{w}_A}{M{w}_B}=0.5\times \frac{2}{3}=\frac{1}{3}$
$\therefore M{w}_A:M{w}_B=1:3$

Therefore, the option is $A$