Question

what is the mole fraction of solute in a 3.67 m aqueous solution?
$x_{solute}=$

Explanation:

Step1: Recall molality definition

Molality ($m$) is moles of solute per kilogram of solvent. A 3.67 m aqueous solution means there are 3.67 moles of solute per 1 kg (1000 g) of water.

Step2: Calculate moles of water

The molar - mass of water ($H_2O$) is $M = 18.015\ g/mol$. The number of moles of water, $n_{water}=\frac{1000\ g}{18.015\ g/mol}\approx55.5\ mol$.

Step3: Calculate mole fraction of solute

The mole fraction of solute ($X_{solute}$) is given by the formula $X_{solute}=\frac{n_{solute}}{n_{solute}+n_{solvent}}$. Here, $n_{solute} = 3.67\ mol$ and $n_{solvent}=n_{water}\approx55.5\ mol$. So, $X_{solute}=\frac{3.67\ mol}{3.67\ mol + 55.5\ mol}=\frac{3.67}{3.67 + 55.5}\approx0.062$.

Answer:

0.062