Calculate the percentage loss of mass of hydrated copper[II] sulphate $[C u S O_{4} .5 H_{2} O]$ when it is completely dehydrated.
$C u S O_{4} .5 H_{2} O \to C u S O_{4} + 5 H_{2} O$
[Atomic weights are $C u = 64, S = 32, O = 16, H = 1$ ]

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Solution

The molecular weight of water $H_{2} O = 2 (1) + 16 = 18$ .
When 1 mole of $C u S O_{4} \cdot 5 H_{2} O$ is heated, 5 moles of water are lost.
The molecular weight of $C u S O_{4} \cdot 5 H_{2} O = 64 + 32 + 4 (16) + 5 (18) = 250$ g/mol
Mass of water lost $= 5 (18) = 90$ g
Percentage loss of mass of $C u S O_{4} \cdot 5 H_{2} O = \frac{90}{250} \times 100 = 36$ %.

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