# "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?

Jun 1, 2017

The molarity of the diluted sulfuric acid solution will be $\text{0.256 mol/L}$, or $\text{0.256 M}$.

#### Explanation:

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

${M}_{1} {V}_{1} = {M}_{2} {V}_{2}$,

where $M$ is molarity in $\text{mol/L}$ and $V$ is the volume of the solution in liters.

The initial volume is given in $\text{mL}$, but it needs to be in liters. Convert $\text{60.0 mL}$ into liters. $\text{1 L = 1000 mL}$

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

Known

${M}_{1} = \text{6.07 M"="6.07 mol/L}$

${V}_{1} = \text{0.0600 L}$

${V}_{2} = \text{1.42 L}$

Unknown

M_2=?

Solution

Rearrange the formula above to isolate ${M}_{2}$. Insert the known data into the equation and solve.

${M}_{2} = \frac{{M}_{1} {V}_{1}}{{V}_{2}}$

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 $\text{0.256 M}$ (rounded to three sig figs)