Question

When $2\, g$ of gas $A$ is introduced into an evacuated flask kept of $25^{\circ} C$, the pressure was found to be $1$ atmosphere. If $3\, g$ of another gas $B$ is then added to the same flask, the pressure becomes $1.5\, atm$. Assuming ideal behavior, the ratio of molecular weights $\left(M_{ A }: M_{ B }\right)$ is

• A. $1: 3$
• B. $3: 1$
• C. $2: 3$
• D. $3: 2$

Answer 1

Answer: (A)

Moles $\propto$ Pressure
i. $\frac{2}{M w_{ A }} \propto 1\, atm$
Pressure of $B=1.5-1=0.5\, atm$
ii. $\frac{3}{M w_{ B }} \propto 0.5 \,atm$
$\frac{3}{M w_{ B }} \times \frac{M w_{ A }}{2}=\frac{0.5}{1}$
$\frac{M w_{ A }}{M w_{ B }}=0.5 \times \frac{2}{3}=\frac{1}{3}$
$\therefore M w_{ A }: M w_{ B }=1: 3$