What is the mole fraction of solute in a 3.89 m aqueous solution?

Dec 13, 2016

$\text{Molality}$ $=$ $\text{Moles of solute"/"Kilograms of solvent}$

$\chi = \frac{3.89 \cdot m o l}{3.89 \cdot m o l + \frac{1000 \cdot g}{18.01 \cdot g \cdot m o {l}^{-} 1}} = 0.067$

Explanation:

And $\chi ,$ $\text{mole fraction}$ $=$ $\text{Moles of solute"/"Moles of solute and moles of solvent}$

We have a $3.89$ $\text{molal}$ solution (at least I think we do); i.e. $\text{molality}$ $=$ $3.89 \cdot m o l \cdot k {g}^{-} 1$.

And thus $\chi = \frac{3.89 \cdot m o l}{3.89 \cdot m o l + \frac{1000 \cdot g}{18.01 \cdot g \cdot m o {l}^{-} 1}} = 0.067$