Question

$8$ mg of $C{H}_4$ gas is removed from $6.02\times {10}^{20}$ molecules of $C{H}_4$, then the volume occupied by $C{H}_4$ gas left at $STP$ is:

• A. 5.6 ml
• B. 11.2 ml
• C. 22.4 ml
• D. 5.6 L

Answer 1

The correct option is B 11.2 ml

$1$ mole has $6.02\times {10}^{23}$ molecules

Moles in $6.02\times {10}^{20}$ molecules $=\frac{6.02\times {10}^{20}}{6.02\times {10}^{23}}={10}^{-3}mol$

Initial mass present $=M\times n=16\times {10}^{-3}g$

Removed mass as gas $=8\times {10}^{-3}g$.

So, remaining mass $=8\times {10}^{-3}g[=(16-8)\times {10}^{-3}]$

So, remaining moles $=\frac{1}{2}\times {10}^{-3}$ moles

$1$ mole has $22.4L$ or $22.4\times {10}^{3}ml$.

In $\frac{1}{2}\times {10}^{-3}$ moles volume of the gas $=11.2ml$