# Question 03a5a

Aug 5, 2017

$\textcolor{red}{\text{8.3 mL}}$ of a 6.0 M solution of HCl to make 100.0 mL of a 0.50 M solution of HCl.

#### Explanation:

This is a dilution problem involving molarity. The formula to use is:

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

where $M$ is molarity and $V$ is volume in liters.

The unit for molarity is $\text{mol solute/L of solution}$, or simply $\text{mol/L}$. So a 2.5 M solution contains $\text{2.5 mol solute/L of solution}$. Therefore, all volumes must be in liters. The final volume can be converted into milliliters, as is the case in this question.

Organize the data:

Known

${M}_{1} = \text{6.0 M HCl}$$=$$\text{6.0 mol/L HCl}$

${M}_{2} = \text{0.50 M HCl}$$=$$\text{0.50 mol/L HCl}$

V_2=100.0 color(red)cancel(color(black)("mL"))xx(1"L")/(1000color(red)cancel(color(black)("mL")))="0.1000 L"

Unknown

${V}_{1}$

Solution

Rearrange the formula to isolate ${V}_{1}$. Insert the data into the equation and solve.

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

V_1=(0.50color(red)cancel(color(black)("M"))xx0.1000"L")/(6.0color(red)cancel(color(black)("M")))="0.0083 L"

(rounded to two significant figures)

Convert liters to milliliters.

0.0083color(red)cancel(color(black)("L"))xx(1000"mL")/(1color(red)cancel(color(black)("L")))="8.3 mL"#