Question

The molecular weights of two ideal gases $A$ and $B$ are respectively $100g/mol$ and $200g/mol$. One gram of $A$ occupies $V$ litre of volume at STP. What is the volume (in litre) occupied by one gram of $B$ at STP?

• A. $\frac{V}{2}$
• B. $V$
• C. $V^2$
• D. $2V$

Answer 1

The correct option is A $\frac{V}{2}$

Given: $\text{Molecular weight of A}=100g/mol$
$\text{Molecular weight of B}=200g/mol$
and $w_A=1g$, $w_B=1g$, $V_A=V$
So, $n_A=\frac{1}{100}$
and $n_B=\frac{1}{200}$
Under identical conditions of temperature and pressure:
$\frac{V_A}{V_B}=\frac{n_A}{n_B}$
$\Rightarrow$ $\frac{V_A}{V_B}=\frac{1/100}{1/200}$
$V_B=\frac{V}{2}$