$23.4\text{ g}$ of $\ce{NaCl}$ are dissolved in $100\text{ g}$ of water. Calculate concentration of salt in the solution.
Here's my work:
Number of moles of salt is $0.4\text{ mol}$
Volume of water and hence of solution is $100\text{ g}\times\dfrac{1\text{ cm}^3}{1\text{ g}}\times \dfrac{1\text{ L}}{1000\text{ cm}^3}= 0.1\text{ L}$
Dividing $\dfrac{0.4\text{ mol}}{0.1\text{ L}} = 4\text{ M}$, although the answer is supposed to be $3.89\text{ M}$.
What is the mistake here?
The mistake here is the assumption that only the water contributes to the volume of the solution. However, this solution is $\dfrac{23.4\text{ g}}{23.4\text{ g}+100\text{ g}}=19.0\%\ \ce{NaCl}$ by mass. The density of this solution is greater than the density of water. The Wikipedia article on brine provides the density of salt-water mixture of varying mass percents. The density of your solution is about $1.14\text{ g/cm}^3$.
Thus, the volume of your solution is:
$123.4\text{ g}\times \dfrac{1\text{ cm}^3 }{1.14\text{ g}}= 108.2\text{ cm}^3=0.1082\text{ L}$
However, using this value gets you $0.370\text{ M}$.
You may not have been given the density of salt solutions. You were probably expected to use the density of pure water, but forgot to include the total mass of the solution and not just the mass of the water. However, using this method, I get $0.324\text{ M}$.
I am not able to calculate the "correct" by any means. If this answer came from a solutions manual, it is possible that the "correct" answer contains an error that the editorial process did not catch.
