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 1L 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

Hint: STP means standard temperature and pressure. They are the standard sets of conditions for experimental measurements to be established to allow comparisons between different sets of data. Further, the ideal gas law can be written in terms of number of molecules of gas, i.e. $PV = NkT$ where P is pressure, T is temperature, V is volume and K is the Boltzmann constant.

Formula used:
Number of molecules$ = $number of moles $ \times $$6.023 \times {10^{23}}$ (Avogadro number)

Complete step by step answer:
In the question, we have been given that, in a mixture of gases the volume content of gas is $0.06\% $ at STP.
So, this means that
Volume content of gas A in a mixture of A and B is $0.06\% $
Now, $1$L of mixture contains $\dfrac{{0.06}}{{100}} \times 1$$ = 6 \times {10^{ - 4}}L$ of A.
As we know that the volume of I mole of gas $ = 22.4L$
So, the number of moles in $6 \times {10^{ - 4}}L$ of gas at STP is equal to,
$ = \dfrac{{6 \times {{10}^{ - 4}}}}{{22.4}}$
Therefore, according to the formula,
Number of molecules $ = $number of moles$ \times 6.023 \times {10^{23}}$
$ = \dfrac{{6 \times {{10}^{ - 4}}}}{{22.4}} \times 6.023 \times {10^{23}}$
$ = 1.61 \times {10^{19}}$ Molecules

Hence, option D is correct.

Note:
The number $6.02214076 \times {10^{23}}$ (the Avogadro number) was chosen so that the mass of one mole of a chemical compound in grams is numerically equal, for most practical purposes, to the average mass of one molecule of the compound in Daltons. A mole corresponds to the mass of a substance that contains $6.023 \times {10^{23}}$ particles of the substance.