Answer to Question #142198 in General Chemistry for jack

Question

1. You have 75.0 cm3 of a 0.150 mol dm-3 solution of zinc sulphate (ZnSO4). What mass of zinc sulphate crystals will be left behind on evaporation of the water?

Answer 1

The concentration of zinc sulphate (ZnSO₄)

= 0.150 mol dm-3

Means,

1 dm3 solution contain 0.150 mol ZnSO₄

Volume of the solution is = 75.0 cm3

As, 1 cm = 10^–1 dm

1 cm3 = (10^–1 dm)3

= 10^–3 dm3

So,

Volume of the solution is = 75.0 cm3

= 75.0 × 10^–3 dm3

So,

1 dm3 solution contain 0.150 mol ZnSO₄

75.0 × 10^–3 dm3 solution contain

= 75.0 × 10^–3 × 0.150 mole

= 11.25×10^–3 mole

So, the given amount of solution contain 11.25×10^–3 mole ZnSO₄.

On evaporation 11.25×10^–3 mole of ZnSO₄ will left behind.

Molar mass of ZnSO₄ = 161.47 g/mole

So,

1 mole ZnSO₄ = 161.47 g/mole ZnSO₄

11.25×10^–3 mole ZnSO₄

= (161.47 × 11.25×10^–3) g

= 1.817 g ZnSO₄

Hence,

1.817 g zinc sulphate (ZnSO₄) crystals will be left behind on evaporation of the water