# How would you prepare 1.0 L of a 0.10 M solution of sulfuric acid from a 3.0 M solution of sulfuric acid?

Jul 9, 2014

This is an example of a dilution calculation. You can find two methods of doing dilution calculations here: How do you do dilution calculations?

${V}_{1}$ = ?; ${c}_{1}$ = 3.0 mol/L
${V}_{2}$ = 1.0 L; ${c}_{2}$ = 0.10 mol/L

Method 1. Using ${V}_{1} {c}_{1} = {V}_{2} {c}_{2}$

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

V_1 = V_2 × c_2/c_1 = 1.0 L × $\left(0.10 \text{mol/L")/(3.0"mol/L}\right)$ = 0.033 L = 33 mL

Method 2. Calculating moles

${V}_{2} {c}_{2}$ = 1.0 L × $\left(0.10 \text{mol")/(1"L}\right)$ = 0.10 mol, so

${V}_{1} {c}_{1}$ = 0.10 mol

${V}_{1} = \left(0.10 \text{mol")/(3.0"mol·L⁻¹}\right)$ = 0.033 L = 33 mL

In each case, you would place 33 mL of the concentrated solution in a 1 L volumetric flask. Then you would add enough distilled water to fill the flask to the 1 L mark