Question

The vapour density of a mixture of gas A (molecular mass = 40 ) and gas B (molecular mass = 80) is 25. Then, mole percentage of gas B in the mixture would be:

• A. 43%
• B. 30%
• C. 25%
• D. 45%

Answer 1

The correct option is C 25%

As vapour density of mixture is 25, molar mass of mixture is $2\times 25=50.$
Now, assume mole % of B in mixture is x.
$\text{Average mass}=\frac{\left(\right(100-x)\times 40+x\times 80)}{100}=50\therefore x=25\%$

Hence, option (a) is correct.