Answer:
Approximately [tex]0.180[/tex].
Explanation:
The mole fraction of a compound in a solution is:
[tex]\displaystyle \frac{\text{Number of moles of compound in question}}{\text{Number of moles of all particles in the solution}}[/tex].
In this question, the mole fraction of [tex]\rm KCl[/tex] in this solution would be:
[tex]\displaystyle X_\mathrm{KCl} = \frac{n(\mathrm{KCl})}{n(\text{All particles in this soluton})}[/tex].
This solution consist of only [tex]\rm KCl[/tex] and water (i.e., [tex]\rm H_2O[/tex].) Hence:
[tex]\begin{aligned} X_\mathrm{KCl} &= \frac{n(\mathrm{KCl})}{n(\text{All particles in this soluton})}\\ &= \frac{n(\mathrm{KCl})}{n(\mathrm{KCl}) + n(\mathrm{H_2O})}\end{aligned}[/tex].
From the question:
Apply the formula [tex]\displaystyle n = \frac{m}{M}[/tex] to find the number of moles of [tex]\rm KCl[/tex] and [tex]\rm H_2O[/tex] in this solution.
[tex]\begin{aligned}n(\mathrm{KCl}) &= \frac{m(\mathrm{KCl})}{M(\mathrm{KCl})} \\ &= \frac{195.0\; \rm g}{74.6\; \rm g \cdot mol^{-1}} \approx 2.61\; \em \rm mol\end{aligned}[/tex].
[tex]\begin{aligned}n(\mathrm{H_2O}) &= \frac{m(\mathrm{H_2O})}{M(\mathrm{H_2O})} \\ &= \frac{215\; \rm g}{18.0\; \rm g \cdot mol^{-1}} \approx 11.9\; \em \rm mol\end{aligned}[/tex].
The molar fraction of [tex]\rm KCl[/tex] in this solution would be:
[tex]\begin{aligned} X_\mathrm{KCl} &= \frac{n(\mathrm{KCl})}{n(\text{All particles in this soluton})}\\ &= \frac{n(\mathrm{KCl})}{n(\mathrm{KCl}) + n(\mathrm{H_2O})} \\ &\approx \frac{2.61 \; \rm mol}{2.61\; \rm mol + 11.9\; \rm mol} \approx 0.180\end{aligned}[/tex].
(Rounded to three significant figures.)
