Q129CP - Chemistry: Molecular Nature Of Matter And Change

Question

The salinity of 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 MgSO₄ are contained in 1.00 kg of the solution?

Step-by-Step Solution

Answer

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

Step 1: Determination of moles of NaCl and MgSO 4

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

Step 2: Determination of mass of NaCl and MgSO 4

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.