# How milliliters of a 9.0 M H_2SO_ solution are needed to make 0.35 L of a 3.5 M solution?

Sep 29, 2016

You will need $\text{140 mL}$ of $\text{9.0 M H"_2"SO"_4}$ to make $\text{350 mL}$ of a $\text{3.5 M}$ solution.

#### Explanation:

The unit for molarity (M) is $\text{moles of solute"/"liters of solution"="mol/L}$.

When diluting a solution, the amount of solute remains constant, but the volume of the solution increases.

The formula for the dilution of a solution is given below:

$\text{M"_1"V"_1"=M"_2"V"_2}$,

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

Known
$\text{M"_1="9.0 M"="9.0 mol/L}$
$\text{M"_2="3.5 M"="3.5 mol/L}$
$\text{V"_2="0.35 L}$

Unknown
$\text{V"_1}$

Solution
Rearrange the dilution formula to isolate $\text{V"_1}$. Substitute the known values into the equation and solve.

"V"_1=("M"_2"V"_2)/("M"_1)

$\text{V"_1=(3.5cancel"M"xx"0.35 L")/(9.0cancel"M")="0.14 L}$ rounded to two significant figures

$\text{V"_1=0.14 cancel"L"xx(1000 "mL")/(1 cancel"L")="140 mL}$ rounded to two significant figures