Question

A student is making a solution of NaCl in water. If the student uses 8.13 grams of NaCl and enough water to make 6.61 liters of solution, what is the molarity of the student's salt solution? Round your answer to the nearest 0.01 and include unitsproperly abbreviated, but NOT substance!

Answer 1

The molarity can be calculated by the equation:

`C=\frac{n_{\text{solute}}}{V_{\text{solution}}}`

Since we know that we have 8.13 grams of NaCl, we can convert this mass to number of moles using the molar mass of NaCl:

`\begin{gathered} M_(NaCl)=(m_(NaCl))/(n_(NaCl)) \\ M_(NaCl)=1\cdot M_(Na)+1\cdot M_(Cl)=1\cdot22.98976928g\/mol+1\cdot35.453g\/mol=58.44276928g\/mol \end{gathered}`

So, solving for the number of moles, we have:

`n_(NaCl)=(m_(NaCl))/(M_(NaCl))=(8.13g)/(58.44276926g\/mol)=0.13911\ldots mol`

We also know the volume of the solution: 6.61 L, so:

`\begin{gathered} n_{\text{solute}}=0.13911\ldots mol \\ V_{\text{solution}}=6.61L \\ C=(0.13911\ldots mol)/(6.61L)=0.021045\ldots mol\/L\approx0.02mol\/L \end{gathered}`

So, the molarity is approximately 0.02 mol/L.