To prepare 250mL of calcium chloride solution with a molar concentration of 1.20mol/L, what mass of calcium chloride would be required?

33.30 grams of CaCl2 will be required

Explanation:

Given,

Volume of solution, V= 250 ml

Molarity of solution, M= 1.20 mol/L

Molecular mass of CaCL2, S= 40+(35.5 X 2)= 111

We know,

Required mass, W= SVM/1000

Now,

W = (111 X 250 X 1.20)/1000

    = 33300/1000

    = 33.30

Therefore, 33.30 grams of Calcium Chloride will be required.