How can we determine the boiling point of 1 molar aqueous solution (density 1.06 g per ml) of KBr? If I use the colligative property formula than I require molality instead of molarity. Here the density given is the one of the solution not the solvent.


The key here is to find the mass of the solvent. If you take one mL of the solution, it will contain $10^{-3} \mathrm{mol}$ of KBr.

The mass of $10^{-3} \mathrm{mol}$ of KBr is $10^{-3}\times(39+79,9)=10^{-3}\times 118.9=0.1189 \mathrm{g}$.

Or, the mass of one mL of the solution is $1.06 \mathrm{g}$. This means that the mass of the solvent in one mL of the solution is $1.06 -0.1189=0.9411 \mathrm{g}=0.9411\times 10^{-3}\mathrm{kg} $.

The molality is: $$m=\frac{10^{-3} \mathrm{mol}}{0.9411\times 10^{-3}\mathrm{kg}}=1.063\,\, \mathrm{mol/kg} $$


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.