Question

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

Answer 1

Answer:

361.5 ml of water will required to make 1.00 L of 0.100 M $Ba(OH)_2$.

Explanation:

Mass of barium hydroxide = 51.24 g

Volume of solution $V$= 1.20 L

Molarity of the solution = $M_1$

$M_1=\frac{51.24 g}{154.33 g/mol\times 1.20 L}=0.2766 M$

Let the volume of the solution 0.2766 M to be diluted be $V_1$

Desired concentration of barium hydroxide solution =$M_2=0.100 M$

Volume of the 0.1000 M solution =$V_2=1.00 L$

$M_1V_1=M_2V_2$

$V_1=\frac{M_2\times V_2}{M_1}=\frac{0.100 M\times 1.00 L}{0.2766 M}$

$V_1=0.3615 L=361.5 mL$

361.5 ml of water will required to make 1.00 L of 0.100 M $Ba(OH)_2$.