Question

`"60.0 mL"` of a `"6.07 M"` stock solution of sulfuric acid is diluted to `"1.42 L"`. What is the molarity of the diluted solution?

Answer 1

The molarity of the diluted sulfuric acid solution will be `"0.256 mol/L"`, or `"0.256 M"`.

Explanation:

When diluting a solution, you can determine the molarity of the diluted solution using the formula:

`M_1V_1=M_2V_2`,

where `M` is molarity in `"mol/L"` and `V` is the volume of the solution in liters.

The initial volume is given in `"mL"`, but it needs to be in liters. Convert `"60.0 mL"` into liters. `"1 L = 1000 mL"`

`60.0color(red)cancel(color(black)("mL"))xx(1"L")/(1000color(red)cancel(color(black)("mL")))="0.0600 L"`

Organize your data.

Known

`M_1="6.07 M"="6.07 mol/L"`

`V_1="0.0600 L"`

`V_2="1.42 L"`

Unknown

`M_2=?`

Solution

Rearrange the formula above to isolate `M_2`. Insert the known data into the equation and solve.

`M_2=(M_1V_1)/(V_2)`

`M_2=((6.07"mol")/("1L")xx0.0600color(red)cancel(color(black)("L")))/(1.42color(red)cancel(color(black)("L")))="0.256 mol/L"` or `"0.256 M"` (rounded to three sig figs)