$23.4\text{ g}$ 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?

  • $\begingroup$ Welcome to Chemistry.SE! To acquaint yourself with this page, take the tour and visit the help center. Furthermore this tutorial shows you how math and chemical formulae can be nicely formatted on this site. $\endgroup$ – Philipp Dec 6 '14 at 11:49
  • $\begingroup$ Are you sure you mean melted? The melting point of sodium chloride is much higher than the boiling point of water. $\endgroup$ – Ben Norris Dec 6 '14 at 15:40
  • $\begingroup$ @BenNorris in some non English languages the words for melting and dissolution are similar. $\endgroup$ – Gimelist Dec 6 '14 at 15:42

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.

  • $\begingroup$ you didn't forget it but yet failed to achieve said result so i am rather confused...begging your pardon... $\endgroup$ – Bak1139 Dec 6 '14 at 16:27
  • $\begingroup$ also density of salt water is not given nor any other density and i did use the density of pure water when calc the general volume of the solution $\endgroup$ – Bak1139 Dec 6 '14 at 16:31
  • 3
    $\begingroup$ I cannot get anything close to the "correct" answer, so I suspect the answer you were given has a mistake in it. $\endgroup$ – Ben Norris Dec 7 '14 at 11:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.