# What is the mole fraction of KCl in an aqueous solution that contains 26.3% KCl?

Sep 7, 2016

$0.079$

#### Explanation:

This question is impossible to answer without making some assumptions.

I am going to assume that we are dealing with a $\text{w/w}$ (weight by weight) percentage solution (for more information on percentage solutions please see this blog post: Calculating Percentage Solutions). The reason I am working on $\text{w/w}$ is because if this is a $\text{w/v}$ solution I would have $26.3$ $g$ of $K C l$ in a final volume of $100$ $m l$, and as we will see, I need to know the exact amount of water added. The difference between a $\text{w/w}$ and $\text{w/v}$ solution is most probably negligible, but science is all about being accurate.

The mole fraction is the number of moles of something divided by the number of total moles present.

$\text{mole fraction" = "moles" / "total moles present}$

First, let us work out the number of moles of $K C l$ present. A 26.3% $\text{w/w}$ solution would contain $26.3$ $g$ of $K C l$ per $100$ $g$ of solution.

The molecular weight of $K C l$ is: $74.5513$ $g$/$m o l$

Hence, $26.3$ $g$ of $K C l$ is $\frac{26.3}{74.5513}$ = $0.353$ moles.

A $\text{w/w}$ percentage solution has a final weight of $100$ $g$. We have $26.3$ $g$ of $K C l$, so the water must weigh $100 - 26.3$ = $73.7$ $g$.

The molecular weight of water is: $18.01528$ $g$/$m o l$.

Hence we have, $\frac{73.7}{18.01528}$ = $4.091$ moles.

Therefore we have:

"mole fraction" = 0.353 / ("("0.353 + 4.091")")

$\text{mole fraction} = \frac{0.353}{4.444}$

$\text{mole fraction} = 0.079$