Question

$2g$ of gas $A$ introduced in a evacuated flask at ${25}^{o}C$. The pressure of the gas is $1$ atm. Now $3g$ of another gas $B$ added to the same flask, the pressure becomes $1.5$ atm. The ratio of molecular mass of $A$ and $B$ is

• A. $\frac{3}{1}$
• B. $\frac{1}{3}$
• C. $\frac{1}{4}$
• D. $\frac{1}{2}$

Answer 1

Answer: (D)

$\frac{1}{2}$

Partial pressure of Gas A= $1$ atm

Let $n_A=\frac{2}{M_A}$

Partial pressure of Gas $B=1.5-1atm=0.5atm$

Moles of $B$ $n_B=\frac{3}{M_B}$

$\Rightarrow \frac{P_1}{N_1}=\frac{P_2}{n_2}\Rightarrow \frac{1}{\frac{2}{M_A}}=\frac{0.5}{\frac{2}{M_B}}$

$\Rightarrow \frac{M_A}{2}=\frac{M_B}{4}\Rightarrow M_A=\frac{M_B}{2}$

Hence, $M_A:M_B=1:2$ .