# What volume (in mL) of 9.91 M sodium chloride is required to produce 100.0 mL of a 0.398M solution?

Oct 6, 2017

$4.02 m L$

#### Explanation:

Recall that molarity and volume can be expressed in the following ratio:

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

So all you have to do is plug in your values:

${V}_{1} \cdot 9.91 M = 100 m L \cdot 0.398 M$

Solve for ${V}_{1}$ to get:

${V}_{1} = \frac{100 m L \cdot 0.398 M}{9.91 M} = 4.016 m L$

With significant figures, you get a final answer of $4.02 m L$.