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$0.53 \mathrm{gm}$ of $\mathrm{Na}_2 \mathrm{CO}_3$ has been dissolved in $100 \mathrm{~m} /$ of a sodium carbonate solution. The normality of the solution will be
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The correct answer is:
$\frac{N}{10}$
$N=\frac{0.53 \times 1000}{53 \times 100} \Rightarrow N=\frac{1}{10}$
So normality of the solution will be $\frac{N}{10}$
So normality of the solution will be $\frac{N}{10}$
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