Question

1 What is the molarity of a solution made by dissolving 4.17 g of magnesium nitrate [Mg(NO₃)₂] in enough water to make 22.0 mL of solution

2 Upon heating 118 g MgSO₄ · 7 H₂O:

(a) how many grams of water can be obtained?

(b) how many grams of anhydrous compound can be obtained?

Answer 1

1. Weight of magnesium nitrate= 4.17 g

We know that, Molarity= No. Of moles of solute/ volume of solution (in litre).

And no. Of moles= mass of solute/ molar mass of solute.

We have, molar mass of magnesium nitrate= 148.3 g/mol.

Therefore, Molarity= (4.17/148.3)/0.22

= 1.27 M.

02. Heating MgSO4.7H2O :

Mass of MgSO4.7H2O = 118 g.

Molar mass = 246.5 g

Now, 246.5 g of hydrated magnesium sulphate contain (7×18) i.e 126 g of H2O.

Therefore, 118 g of hydrated magnesium sulphate contain :

= (118/246.5)× 126

= 60.3 g of H20.

Therefore, mass of anhydrous compound= 118g - mass of H20

= 118- 60.31 g

= 57.69 g.