Question

Calculate (he molarity of each of the following solutions:

1. 30 g of CO(NO₃)₂ . 6H₂O = in 4.3 L of solution 

2. 30 mL of 0.5 M H₂SO₄ diluted to 500 mL.

Answer 1

1. Molar mass of CO(NO₃)₂ . 6H₂O = 58.7 + 2(14 + 48) + (6 x 8)g mol^-1 

= 58.7 + 124 + 108 g mol^-1 = 290.7 g mol^-1

Volume of solution = 4.3 L

2. 1000 mL of 0.5 M H₂SO₄ Contain H₂SO₄ = 0.5 moles 

30 mL of 0.5 M H₂SO₄ contain H₂SO₄ =  \(\frac{0.5}{100}\) x 30 mole 

= 0.015 mole

Volume of solution = 500 mL = 0.500 L