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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

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Correct option: (c) 2.4 x 1021

Explanation:

Given that 

C2H2O4. 2H2O = 252 mg

One mole molecule of oxalic acid contains 2 mole of H2

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  

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