Question

A solution was prepared by dissolving 195.0 g of KCl in 215 g of water. Calculate the mole fraction of KCl. (The formula weight of KCl is 74.6 g/mol. The formula weight of water is 18.0 g/mol.)

Answer 1

Approximately
`0.180`.

Step-by-step explanation:

The mole fraction of a compound in a solution is:

`\displaystyle \frac{\text{Number of moles of compound in question}}{\text{Number of moles of all particles in the solution}}`.

In this question, the mole fraction of
`\rm KCl` in this solution would be:

`\displaystyle X_\mathrm{KCl} = \frac{n(\mathrm{KCl})}{n(\text{All particles in this soluton})}`.

This solution consist of only
`\rm KCl` and water (i.e.,
`\rm H_2O`.) Hence:

`\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}`.

From the question:

• Mass of
  `\rm KCl`:
  `m(\mathrm{KCl}) = 195.0\; \rm g`.
• Molar mass of
  `\rm KCl`:
  `M(\mathrm{KCl}) = 74.6\; \rm g \cdot mol^(-1)`.

• Mass of
  `\rm H_2O`:
  `m(\mathrm{H_2O}) = 215\; \rm g`.
• Molar mass of
  `\rm H_2O`:
  `M(\mathrm{H_2O}) = 18.0\; \rm g\cdot mol^(-1)`.

Apply the formula
`\displaystyle n = (m)/(M)` to find the number of moles of
`\rm KCl` and
`\rm H_2O` in this solution.

`\begin{aligned}n(\mathrm{KCl}) &= \frac{m(\mathrm{KCl})}{M(\mathrm{KCl})} \\ &= (195.0\; \rm g)/(74.6\; \rm g \cdot mol^(-1)) \approx 2.61\; \em \rm mol\end{aligned}`.

`\begin{aligned}n(\mathrm{H_2O}) &= \frac{m(\mathrm{H_2O})}{M(\mathrm{H_2O})} \\ &= (215\; \rm g)/(18.0\; \rm g \cdot mol^(-1)) \approx 11.9\; \em \rm mol\end{aligned}`.

The molar fraction of
`\rm KCl` in this solution would be:

`\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 (2.61 \; \rm mol)/(2.61\; \rm mol + 11.9\; \rm mol) \approx 0.180\end{aligned}`.

(Rounded to three significant figures.)