What is the mole fraction of 1.0 g of a compound of relative molecular mass 100.0 dissolved in 30.0 g of solvent of relative molecular mass 50.0? - AQA - A-Level Chemistry - Question 9 - 2021 - Paper 3

Similarly, for the solvent:

$n_{solvent} = \frac{m}{M}$

where:

• $m$ is the mass of the solvent (30.0 g)
• $M$ is the molecular mass of the solvent (50.0 g/mol)

Thus,

$n_{solvent} = \frac{30.0 \text{ g}}{50.0 \text{ g/mol}} = 0.6 \text{ mol}$

The mole fraction ($X$) of the compound is calculated using the formula:

$X_{compound} = \frac{n_{compound}}{n_{compound} + n_{solvent}}$

Substituting the values:

$X_{compound} = \frac{0.01 \text{ mol}}{0.01 \text{ mol} + 0.6 \text{ mol}} = \frac{0.01}{0.61} \approx 0.016393$

So, the mole fraction is approximately $1.6 \times 10^{-2}$.

The mole fraction of the compound is approximately $1.6 \times 10^{-2}$, which corresponds to option B.