Question

An industrial sewage water was found to be contaminated with mercuric sulphate. The level of contamination was 34 ppm (by mass). Hg= 200.5,s=32,O=16.

a)Express this in percent by mass

b) Determine the molality of mercuric sulphate in this sample.

Answer 1

34 ppm solution means 34 gram of solute is present in 10⁶ gram of solution.

(i) Percentage by mass

Mass of solution = 10⁶ g
Mass of solute = 34 g
$$\begin{gathered} \mathrm{Percentage}\mathrm{by}\mathrm{mass}=\frac{\mathrm{Mass}\mathrm{of}\mathrm{solute}}{\mathrm{Mass}\mathrm{of}\mathrm{solution}}\times 100 \\ \mathrm{Percentage}\mathrm{by}\mathrm{mass}=\frac{34}{{10}^6}\times 100 \\ \mathrm{Percentage}\mathrm{by}\mathrm{mass}=34\times {10}^{-4}\% \end{gathered}$$

(ii) Molality of solution
$$\begin{gathered} \mathrm{Molar}\mathrm{mass}\mathrm{of}\mathrm{solute}\left({\mathrm{HgSO}}_4\right)=200.5+32+16\times 4=296.5\mathrm{g}/\mathrm{mol} \\ \mathrm{Molality}=\frac{\mathrm{Mass}\mathrm{of}\mathrm{solute}}{\mathrm{Molar}\mathrm{mass}\mathrm{of}\mathrm{solute}}\times \frac{1000}{\mathrm{Mass}\mathrm{of}\mathrm{solvent}\left(\mathrm{grams}\right)} \\ \mathrm{Molality}=\frac{34}{296.5}\times \frac{1000}{{10}^6}(\mathrm{mass}\mathrm{of}\mathrm{solute}\mathrm{is}\mathrm{negligible}\mathrm{as}\mathrm{compared}\mathrm{to}\mathrm{mass}\mathrm{of}\mathrm{solution},\mathrm{thus}\mathrm{mass}\mathrm{of}\mathrm{solution}~\mathrm{mass}\mathrm{of}\mathrm{solvent}) \\ \mathbf{Molality}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}\mathbf{.}\mathbf{147}\mathbf{}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{}\mathbf{m} \end{gathered}$$