Question

what would be the amount of pure CaOH required to prepare 250ml of 0.1 N CaOH solution
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Answer 1

Molecular mass of Ca(OH)2 = 40 + 2×17 = 74

Thus, eq wt = 74/2 = 37

Now, let the volume of solution be 1 litre.

1 eq of Ca(OH)2 weighs = 37g

So, 0.1 eq of Ca(OH)2 weighs = 3.7g

Now, Normality is defined as no. of equivalents per litre.

So,

1 litre of water should contain = 3.7g Ca(OH)2

Therefore, 250mL water should contain = 3.7÷1000×250 = 0.925g Ca(OH)2

Thus, 0.925g of Ca(OH)2 should be dissolved in 250mL water to give a 0.1N solution.

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