What is the mole fraction of $K C l$ $KCl$ in an aqueous solution that contains 26.3% $K C l$ $KCl$ ?

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What is the mole fraction of $K C l$ $KCl$ in an aqueous solution that contains 26.3% $K C l$ $KCl$ ?

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Answer 1 · 2016-09-07T07:33:44

$0.079$ $0.079$

Explanation:

This question is impossible to answer without making some assumptions.

I am going to assume that we are dealing with a $w/w$ $"w/w"$ (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 $w/w$ $"w/w"$ is because if this is a $w/v$ $"w/v"$ solution I would have $26.3$ $26.3$ $g$ $g$ of $K C l$ $KCl$ in a final volume of $100$ $100$ $m l$ $ml$ , and as we will see, I need to know the exact amount of water added. The difference between a $w/w$ $"w/w"$ and $w/v$ $"w/v"$ 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.

$mole fraction=molestotal moles present$ $"mole fraction" = "moles" / "total moles present"$

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

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

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

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

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

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

Therefore we have:

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

$mole fraction=0.3534.444$ $"mole fraction" = 0.353 / 4.444$

$mole fraction=0.079$ $"mole fraction" = 0.079$

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What is the mole fraction of $K C l$ $KCl$ in an aqueous solution that contains 26.3% $K C l$ $KCl$ ?

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