Question

what volume of 0.500 mol/L MgCl2 solution is needed to prepare 250 ml of a solution that has a chloride ion concentration of 0.100 mol/L

Answer 1

When calculating concentration, we always need to remember that the number of moles remain constant.

The formula for concentration is:

`c=(n)/(V)`

So, if we don't add or remove solute, n will remain constant, while c and V vary.

If we want a solution of chloride ion concentration of 0.100 mol/L and 250 mL of volume, we need to first convert 250 mL to L:

`250mL=(250)/(1000)L=0.250L`

Now, we can use the formula for concentration to calculate the number of moles:

`\begin{gathered} c_(Cl^(-))=(n_(Cl^(-)))/(V_(Cl^(-))) \\ n_(Cl^(-))=c_(Cl^(-))V_(Cl^(-)) \\ n_(Cl^(-))=0.100mol/L\cdot0.250L=0.0250mol \end{gathered}`

Now, the dissociation of MgCl₂ will be:

`MgCl_2\to Mg^(2+)+2Cl^-`

So, for each mol of MgCl₂, we will have 2 moles of Cl⁻.

So, if we need 0.0250 mol of Cl⁻, we need half as many of MgCl₂, that is:

`n_{MgCl_(2)}=(1)/(2)n_(Cl^-)=(1)/(2)\cdot0.0250mol=0.0125mol`

Now, we need to calculate the volume necessary of the solution of 0.500 mol/L MgCl₂:

`\begin{gathered} c_{MgCl_(2)}=\frac{n_(MgCl_2)}{V_{MgCl_(2)}} \\ V_(MgCl_2)=\frac{n_(MgCl_2)}{c_{MgCl_(2)}}=(0.0125mol)/(0.500mol/L)=0.0250L=0.0250\cdot1000mL=25mL \end{gathered}`

So, we will need 25 mL os the solution.