Question

5) A 43.3 mL of 7.69 M Sn(NO3)2 is diluted to 137.6mL with water. What is the molarity of the resulting solution?

Answer 1

The molarity or molar concentration of a solution is given by the number of moles of solute divided by the volume of solution:

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

No solute was added when the solution was diluted, so the only terms that change are the concentration and the volume.

So, let 1 denote the situation before the dilution and 2 denote the situation after the dilution.

At first, we have concentration 1, volume 1 and number of moles 1:

`C_1=(n_1)/(V_1)_{}`

After diluting, we have concentration 2, volume 2 and number of moles 2:

`C_2=(n_2)/(V_2)`

However, since no solute was added or removed, the number of moles before and after are the same:

`n_1=n_2`

So, if we solve both equations for n, we have:

`\begin{gathered} C_1=(n_1)/(V_1)\Longrightarrow n_1=C_1V_1 \\ C_2=(n_2)/(V_2)\Longrightarrow n_2=c_2V_2 \end{gathered}`

So:

`\begin{gathered} n_1=n_2 \\ C_1V_1=C_2V_2 \end{gathered}`

The initial concentration and volume are:

`\begin{gathered} C_1=7.69M \\ V_1=43.3mL \end{gathered}`

And it was diluted to 137.6 mL, so this is the final volume:

`V_2=137.6mL`

Solving the equation we have to C2, we have:

`\begin{gathered} C_2=(C_1V_1)/(V_2) \\ C_2=(7.69M\cdot43.3mL)/(137.6mL)=(7.69\cdot43.3)/(137.6)M=2.41989\ldots M\approx2.42M \end{gathered}`

So, the resulting molarity is approximately 2.42 M.