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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.

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

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