Question

From $160$g sample of $S{O}_2,1.2046\times {10}^{24}$ molecules are removed, find the volume of remaining $S{O}_2$ at STP.

• A. $11.2$ litre
• B. $12.2$ litre
• C. $5.2$ litre
• D. $15.4$ litre

Answer 1

Answer: (A)

$11.2$ litre

$1.2046\times {10}^{24}$ molecules of $S{O}_2$$=\frac{1.2046\times {10}^{24}}{6.023\times {10}^{23}}$ $=2$ mole of $S{O}_2$

160 gram $S{O}_2$ contains 2.5 mole $S{O}_2$ and 2 moles are removed

So, 0.5 moles are remaining.

Volume of one mole of $S{O}_2$ at STP is $22.4$ litre

Therefore volume of 0.5 mole of $S{O}_2$ is 22.4/2 $=11.2$ litre