Question

If 15 mg of $N_2{O}_3$ is added to $4.82\times {10}^{20}$ molecules of $N_2{O}_3$the total volume occupied by the gas at STP is:

• A. 0.044 L
• B. 0.022 L
• C. 0.22 L
• D. 0.44 L

Answer 1

The correct option is B 0.022 L

Gram molecular weight of $N_2{O}_3$ is 76 $g$.

76 $g$ of $N_2{O}_3$ contains - $6.023\times {10}^{23}$ molecules.

$\therefore$ Weight of $4.82\times {10}^{20}$ molecules$=\frac{76\times 4.82\times {10}^{20}}{6.023\times {10}^{23}}g=0.061g$

$15$ mg of $N_2{O}_3=0.015g$

$\therefore$ Total weight of $N_2{O}_3=0.061+0.015=0.076g$

$76$ g occupies 22.4 $L$ at STP.

0.076 $g$ occupies at $STP$ $=\frac{0.076\times 22.4}{76}=0.022L$

Hence, the correct option is $\text{B}$