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$1.25 \mathrm{~g}$ of a sample of $\mathrm{Na}_{2} \mathrm{CO}_{3}$ and $\mathrm{Na}_{2} \mathrm{SO}_{4}$ is
dissolved in $250 \mathrm{ml}$ solution. $25 \mathrm{ml}$ of this solution neutralises $20 \mathrm{ml}$ of $0.1 \mathrm{~N} \mathrm{H}_{2} \mathrm{SO}_{4}$. The $\%$ of $\mathrm{Na}_{2} \mathrm{CO}_{3}$
in this sample is
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dissolved in $250 \mathrm{ml}$ solution. $25 \mathrm{ml}$ of this solution neutralises $20 \mathrm{ml}$ of $0.1 \mathrm{~N} \mathrm{H}_{2} \mathrm{SO}_{4}$. The $\%$ of $\mathrm{Na}_{2} \mathrm{CO}_{3}$
in this sample is
Solution:
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Verified Answer
The correct answer is:
$84.8 \%$
Let the amount of $\mathrm{Na}_{2} \mathrm{CO}_{3}$ present in the mixture be $\mathrm{xg} . \mathrm{Na}_{2} \mathrm{SO}_{4}$ will not react with $\mathrm{H}_{2} \mathrm{SO}_{4} .$ Then
$\frac{x}{53}=\frac{20 \times 0.1 \times 10}{1000} \therefore x=1.06 g$
$\therefore$ Percentage of $\mathrm{Na}_{2} \mathrm{CO}_{3}=$
$\frac{1.06 \times 100}{1.25}=84.8 \%$
$\frac{x}{53}=\frac{20 \times 0.1 \times 10}{1000} \therefore x=1.06 g$
$\therefore$ Percentage of $\mathrm{Na}_{2} \mathrm{CO}_{3}=$
$\frac{1.06 \times 100}{1.25}=84.8 \%$
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