Question

5g potassium sulphate is dissolved in 250 ml solution. What volume of this solution will be

sufficient to precipitate 1.2 g barium sulphate from barium chloride solution (Ba=137)

25. 1.42 g of a mixture of calcium carbonate and magnesium carbonate was heated to a constant​

Answer 1

Answer:

I think the second part of this question is not  completed.

For the first part:

According to the problem:

K2SO4 + BaCl2 → BaSO4 + 2KCl

Molarity of K2SO4 solution =5174.25×1000250=0.11 M

1 mol of BaSO4 requires 1 mol of K2SO4

So, 233.38 g of BaSO4 requires 174.25 g of K2SO4

1.2 g of BaSO4 requires 174.25/233.38×1.2=0.895 g of K2SO4

or, Moles of K2SO4 required is 0.895174.25=0.005 mol

We know that, M1V1=M2V2

or, 0.11×V1=0.005

or, V1=0.0050.11=0.045 L

Hence 45 ml of this solution will be sufficient.