Question

$1.2475$ g of crystalline copper sulphate was dissolved in water and excess of KI was added. The liberated iodine consumed $50$ mL $\frac{N}{10}N{a}_2{S}_2{O}_3$ solution to reach the end point of the titration.The number of water molecules of hydration in crystalline copper sulphate salt is :

Answer 1

$4I^{-}+2C{u}^{2+}\to C{u}_2{I}_2\downarrow +I_2$

$I_2+2S_2{O}_3^{2-}\to 2I^{-}+S_4{O}_6^{2-}$

Thus $2$ moles of hydrated copper sulphate corresponds to $1$ mole of iodine and $1$ mole of iodine corresponds to $2$ moles of $N{a}_2{S}_2{O}_3$. Hence, $2$ mole of hydrated copper sulphate corresponds to $2$ moles of $N{a}_2{S}_2{O}_3$.

Number of moles of hydrated copper sulphate $=1\times \frac{50}{1000}\times \frac{1}{10}=0.005$ moles.

The mass of hydrated copper sulphate is $1.2475$ g.

The molar mass of hydrated copper sulphate is $\frac{1.2475}{0.005}=249.5$ g/mol.

The molar mass of anhydrous copper sulphate is $159.5$ g/mol.
The molar mass of water is $18$ g/mol.

The number of water molecules of hydration in crystalline copper sulphate salt is $\frac{249.5-159.5}{18}=\frac{90}{18}=5$.