A51.35 g sample of ba(oh)2 (molar mass 171.32 g) is dissolved in enough water to make 1.20 liters of solution. how many ml of this solution must be diluted with water in order to make 1.00 liters of 0.100 m ba(oh)2?

We must diluted 0.40 L in order to make 1 L of 0.1 M Ba(OH)₂

Explanation:

There is something called dilution factor which can help to do this.

Dilution factor is the relationship between diluted and concentrated concentrations, or the quotient  between the volume of the concentrated solution and the volume of the diluted. It also applies to molarity (diluited or concentrated).

Molarity in main solution is: mol/L

Moles = mass / molar mass

51.35 g / 171.32 g/m = 0.299 moles

This moles are in 1.20 L, so in 1 L we have:

0.299 moles/ 1.20 L = 0.249 M

This is the concentrated solution and molarity of diluited is data, 0.1 M.

Dilution factor : diluited/concentrated

0.1 M / 0.249 M = 0.40

Conc. vol. / Diluited vol. = 0.40

Conc. vol / 1 L = 0.40

1 L . 0.40 = Conc. vol = 0.40 L

what is the question?

1 is the answer to this problem