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QUESTION

# A 51.24 g sample of Ba(OH)2 is dissolved in enough water to make 1.20 L of solution. How many mL of this solution must be diluted with water to make 1.00 L, of 0.100 M Ba(OH)2?

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

Step 1. Calculate the of the stock solution.

Moles of Ba(OH)₂ = "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 = "moles"/"litres" = "0.299 05 mol"/"1.20 L" = 0.2492 mol/L

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

See

c_1V_1 = c_2V_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 = "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.