Question

In a mixture of gases, the volume content of a gas is $0.06\%$ at STP. Calculate the number of molecules of the gas in 1 L of the mixture.

• A. $1.613\times {10}^{23}$
• B. $6.023\times {10}^{23}$
• C. $1.61\times {10}^{27}$
• D. $1.61\times {10}^{19}$

Answer 1

The correct option is D $1.61\times {10}^{19}$

Volume content of gas A in a mixture of A and B =0.06%

1 L of mixture contains= $\frac{0.06}{100}\times 1=6\times {10}^{-4}L$ of A

At STP, 1 mol of gas has volume =22.4 L

number of moles in $6\times {10}^{-4}$ L gas at STP=$\frac{6\times {10}^{-4}}{22.4}$

number of molecules=number of moles $\times 6.023\times {10}^{23}$

$=\frac{6\times {10}^{-4}}{22.4}\times 6.023\times {10}^{23}$

$=1.61\times {10}^{19}$ molecules