A gas mixture is made by combining 7.0 g each of Ar, Ne, and an unknown diatomic gas. At STP, the mixture occupies a volume of 16.60 L. What is the molar mass of the unknown gas?

Nov 17, 2016

Considering molar mass of

$A r \to 40 \text{ g/mol}$

$N e \to 20 \text{ g/mol}$

$\text{Molar mass of unknown gas"->m" g/mol}$

The total number of moles Ar , Ne and unknown gas in the gas mixture each having mass of 7g is

$= \left(\frac{7}{40} + \frac{7}{20} + \frac{7}{m}\right) m o l$

Total Volume of these three gases at STP is

$\left(\frac{7}{40} + \frac{7}{20} + \frac{7}{m}\right) m o l \times 22.4 \frac{L}{\text{mol}} = 16.60 L$

$\implies \left(\frac{7}{40} + \frac{7}{20}\right) 22.4 L + \frac{7}{m} \times 22.4 L = 16.60 L$

$\implies \frac{21}{40} \times 22.4 + \frac{7}{m} \times 22.4 = 16.60$

$\implies 11.76 + \frac{7}{m} \times 22.4 = 16.60$

$\implies \frac{7}{m} \times 22.4 = \left(16.60 - 11.76\right) = 4.24$

$\implies m = \frac{7 \times 22.4}{4.24} \approx 36.98 \text{g/mol}$