Question

The salinityof a solution is defined as the grams of total salts per kilogram of solution. An agricultural chemist uses a solution whose salinity is 35.0 g/kg to test the effect of irrigating farmland with high-salinity river water. The two solutes are NaCl and MgSO₄, and there are twice as many moles of NaCl as MgSO₄. What masses of NaCl and MgSO₄are contained in 1.00 kg of the solution?

Short answer

The mass of NaCl and MgSO₄ in the solution is 17.2 g and 17.8 g, respectively.

Step-by-step solution

Determination of moles of NaCl and MgSO4

Let the moles of MgSO₄ in 1.00 kg solution be ‘a’ and the moles of NaCl in 1.00 kg solution be ‘2a’.

Conversion of moles into mass,

$$\begin{gathered} \mathbf{Mass}\mathbf{}\mathbf{of}\mathbf{}\mathbf{Mg}\mathbf{}{\mathbf{SO}}_{\mathbf{4}}{\mathbf{=}}^{\mathbf{\text{'}}}{\mathbf{a}}^{\mathbf{\text{'}}}\mathbf{mol}\mathbf{\times }\mathbf{120}\mathbf{.}\mathbf{38}\mathbf{g}\mathbf{/}\mathbf{mol} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{Mass}\mathbf{}\mathbf{of}\mathbf{}\mathbf{NaCI}{\mathbf{=}}^{\mathbf{\text{'}}}\mathbf{2}{\mathbf{a}}^{\mathbf{\text{'}}}\mathbf{mol}\mathbf{\times }\mathbf{58}\mathbf{.}\mathbf{44}\mathbf{g}\mathbf{/}\mathbf{mol} \end{gathered}$$

The salinity of the solution is 35.0 g/kg.

So, mass of solute in 1.00 kg = 35.0 g.

Now, moles to the given mass are,

$$\begin{gathered} \left({}^{\text{'}}{a}^{\text{'}}\mathrm{mol}\times 120.38g/\mathrm{mol}\right)\mathbf{\times }\left({}^{\text{'}}2a^{\text{'}}\mathrm{mol}\times 58.44g/\mathrm{mol}\right)\mathbf{=}\mathbf{35}\mathbf{.}\mathbf{0}\mathbf{g} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{a}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{1475}\mathbf{mol} \end{gathered}$$

Thus,

Moles of MgSO₄ = ‘a’ = 0.1475 mol

Moles of NaCl = ‘2a’ = 0.295 mol

Determination of mass of NaCl and MgSO4

The mass of NaCl is,

$$\begin{gathered} \mathbf{Mass}\mathbf{=}\left({}^{\text{'}}2a^{\text{'}}\mathrm{mol}\times 58.44g/\mathrm{mol}\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{295}\mathbf{mol}\mathbf{\times }\mathbf{58}\mathbf{.}\mathbf{44}\mathbf{g}\mathbf{/}\mathbf{mol} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{17}\mathbf{.}\mathbf{2}\mathbf{g} \end{gathered}$$

The mass of MgSO₄ is,

$$\begin{gathered} \mathbf{Mass}\mathbf{=}{\mathbf{(}}^{\mathbf{\text{'}}}\mathbf{2}{\mathbf{a}}^{\mathbf{\text{'}}}\mathbf{mol}\mathbf{\times }\mathbf{120}\mathbf{.}\mathbf{38}\mathbf{g}\mathbf{/}\mathbf{mol}\mathbf{)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{1475}\mathbf{mol}\mathbf{\times }\mathbf{120}\mathbf{.}\mathbf{38}\mathbf{g}\mathbf{/}\mathbf{mol} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{17}\mathbf{.}\mathbf{8}\mathbf{g} \end{gathered}$$

Thus, the mass of NaCl and MgSO₄ in the solution is 17.2 g and 17.8 g, respectively.