Question aa595

Jul 1, 2017

${\chi}_{\text{KCl}} = 0.0794$

Explanation:

We're asked to find the mole fraction of $\text{KCl}$ given that is present in a mass percentage of 26.3%.

This means that in a $100$-$\text{g}$ sample of solution, there are $26.3$ $\text{g KCl}$ and $73.7$ $\text{g H"_2"O}$.

Let's convert these mass values to grams using the molar mass of each substance:

26.3cancel("g KCl")((1color(white)(l)"mol KCl")/(74.55cancel("g KCl"))) = 0.353 $\text{mol KCl}$

73.7cancel("g H"_2"O")((1color(white)(l)"mol H"_2"O")/(18.02cancel("g H"_2"O"))) = 4.09 $\text{mol H"_2"O}$

The mole fraction of $\text{KCl}$ is

chi_"KCl" = "mol KCl"/"total moles" = (0.353color(white)(l)"mol KCl")/(0.353color(white)(l)"mol KCl" + 4.09color(white)(l)"mol H"_2"O")

= color(red)(0.0794#