# A stock solution contains 51.24 g of barium hydroxide in 1.20 L of solution. How do I make 1.00 L of 0.100 mol/L barium hydroxide from the stock solution?

Sep 15, 2014

You must dilute 401 mL with enough water to make 1.00 L of solution.

Step 1. Calculate the molarity of the stock solution.

Moles of Ba(OH)₂ = ${\text{51.24 g Ba(OH)"_2 × ("1 mol Ba(OH)"_2)/("171.34 g Ba(OH)"_2) = "0.299 05 mol Ba(OH)}}_{2}$

Molarity = $\text{moles"/"litres" = "0.299 05 mol"/"1.20 L}$ = 0.2492 mol/L

Step 2. Calculate the volume of stock solution that is needed.

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

${c}_{1}$ = 0.2492 mol/L; ${V}_{1}$ = ?
${c}_{2}$ = 0.100 mol/L; ${V}_{2}$ = 1.00 L

V_1 = V_2 × c_2/c_1 = $\text{1.00 L" × "0.100 mol/L"/"0.2492 mol/L}$ = 0.401 L = 401 mL

So, to make your solution, you would dilute 401 mL of stock solution with enough water to make 1.00 L.