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1 g of the carbonate of a metal was dissolved in 25 mL of N-HCI. The resulting liquid required 5 mL of N-NaOH for neutralisation. The Ew of the metal carbonate is

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Detailed Solution

$HCl \Rightarrow 25 \times 1 = 25 m Eq Excess HCl \Rightarrow 5 \times 1 = 5 m Eq HCl used \Rightarrow 25 - 5 = 20 m Eq ∴ m Eq of HCl = m Eq of carbonate$

$\begin{array}{l} \frac{20}{1000} Eq = \frac{Weight}{Ew} of carbonate = \frac{1}{Ew} \\ ∴ Ew = \frac{1000}{20} = 50 \end{array}$

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