Question

How many molecules of water of hydration are present in 252 mg of oxalic acid (C₂H₂O₄.2H₂O) ?

(a) 1.2 x 10²¹ 

(b)1.7 × 10²¹ 

(c) 2.4 x 10²¹ 

(d) 3.4 x 10²¹

Answer 1

Correct option: (c) 2.4 x 10²¹

Explanation:

Given that 

C₂H₂O₄. 2H₂O = 252 mg

One mole molecule of oxalic acid contains 2 mole of H₂O 

So in 2 × 10^–3 moles of oxalic acid amount of water will be = 2 × 2 × 10^–3 = 4 × 10^–3 mole 

So, No. of molecules of water of hydration = 4 × 10^–3 × 6.023 × 10²³ 

= 24.08 × 10²⁰ = 2.4 × 10²¹