Question

A 74.28-g sample of Ba(OH)2 is dissolved in enough water to make 2.450 liters of solution.

How many mL of this solution must be diluted with water in order to make 1.000 L of 0.100 M Ba(OH)2?

Answer 1

First we need to determine the number of moles of Ba(OH)₂ in the initial solution:

$n(Ba(OH)_2)=\frac{74.28g}{171.34g/mol}=0.4335mol$

In order to prepare 1.000 L of 0.100M Ba(OH)₂ solution, 0.100 moles of Ba(OH)₂ are required.

Therefore, the volume of the initial solution that must be diluted equals:

$V=2.450L\times\frac{0.100mol}{0.4335mol}=0.565L=565mL$

Answer: 565 mL