# For a solution with benzene as the solvent, what is the molality (with respect to the solute) if the solution has a solute mol fraction of 0.2?

Apr 19, 2017

Since the mol fraction of solute is ${\chi}_{i} = 0.2$, this indicates that we have:

${\chi}_{i} = 0.2 = \frac{{n}_{i}}{{n}_{i} + {n}_{j}}$

$= \left(\text{0.2 mols solute")/("0.2 mols solute" + "0.8 mols solvent}\right)$

where ${n}_{i}$ is the mols of $i$.

Recall that molality is:

$\text{mols solute"/"kg solvent}$

Fortunately, we know that the solvent is benzene, so the molar mass is:

${M}_{j} = 6 \times 12.011 \text{g/mol C" + 6 xx 1.0079 "g/mol H}$

$=$ $\text{78.1134 g/mol}$

This tells us that the molality is:

color(blue)(m_"soln") = "0.2 mols solute"/(0.8 cancel"mols benzene") xx (cancel"1 mol benzene")/(78.1134 cancel"g benzene") xx (1000 cancel"g")/"1 kg"

$=$ $\textcolor{b l u e}{\text{3.20 molal}}$