Question

The weight of $181.60\text{mL}$ of a diatomic gas at $0{}^{\circ }\text{C}$ and $1\text{bar}$ pressure is $1\text{g}$. The mass of one molecule (in $\text{g}$) of the gas is:
$(N_A$ is the Avogadro's constant)

• A. $\frac{62.5}{N_A}$
• B. $\frac{125}{N_A}$
• C. $\frac{120}{N_A}$
• D. $\frac{75.5}{N_A}$

Answer 1

The correct option is B $\frac{125}{N_A}$

We know,
At STP, one mole of any gas occupies $22.7\text{L}$.
Thus, $181.60\text{mL}\text{of the diatomic gas}=\frac{181.60}{22700}\text{mol}=0.008\text{mol}$
So, according to the question,
Mass of $0.008\text{mol}$ of the gas $=1\text{g}$
$\text{Hence, mass of 1 mol of the gas}=\frac{1}{0.008}\text{g}=125\text{g}$
$\text{So, the mass of one molecule of the diatomic gas}=\frac{125}{N_A}$