Question

What volume of 0.320 mol/L silver nitrate solution would contain 0.240 mol of silver nitrate?

Silver nitrate is used extensively in making photographic plates and films. What volume of 0.320 mol/L silver nitrate solution would contain 0.240 mol of silver nitrate?

Answer 1

You would have to add `"0.33 L"` `("330 mL")` of solvent to the `"0.320 M"` solution in order to dilute it to a `"0.240 M"` solution. The resulting volume of the `"0.240 M"` solution would be `"1.33 L"`.

Explanation:

This question is poorly worded. There is no volume of a `"0.320 mol/L"` solution that would contain `"0.240 mol/L"` silver nitrate. By definition, a solution is thoroughly mixed and the components are uniformly distributed. I think what is meant by this question, is what volume is needed to dilute the `"0.320 mol/L"` solution to `"0.240 mol/L"`.

`"0.320 mol/L"` `=` `"0.320 M"`

`"0.240 mol/L"` `=` `"0.240 M"`

Use the dilution equation:

`M_1V_1=M_2V_2`,

where:

`M_1` and `V_1` are the initial concentration and volume, and `M_2` and `V_2` are the final concentration and volume.

Since no initial volume is given, I'm going to use `"1.00 L"`. In order for the `"0.320 M"` to be diluted to a `"0.240 M"` solution, a measure of solvent must be added in order to make the dilution.

Known

`M_1="0.320 mol/L"="0.320 M"`

`V_1="1.00 L"`

`M_2="0.240 mol/L"="0.240 M"`

Unknown

`V_2`

Solution

Rearrange the dilution equation to isolate `V_2`. Plug in the known values and solve.

`V_2=(M_1V_1)/M_2`

`V_2=(0.320"M"xx1.00"L")/(0.240"M")` `=` `1.3bar3` `"L"`

You would have to add `"0.33 L"` `("330 mL")` of solvent to the `"0.320 M"` solution in order to dilute it to a `"0.240 M"` solution. The resulting volume of the `"0.240 M"` solution would be `"1.33 L"`.