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A 300 mL solution is made by dissolving 8g sample of \mathrm{Na}_2 \mathrm{CO}_3 and \mathrm{Na}_2 \mathrm{SO}_4 only. 20 mL of 0.1 \mathrm{H}_2 \mathrm{SO}_4 is neutralised by 25 mL of the above solution. What is the % of \mathrm{Na}_2 \mathrm{SO}_4 taken?

Option: 1

58.3 %


Option: 2

68.2 %


Option: 3

88.8 %


Option: 4

90 %


Answers (1)

best_answer

\mathrm{Na}_2 \mathrm{CO}_3 reacts with \mathrm{H}_2 \mathrm{SO}_4 as.

\mathrm{Na}_2 \mathrm{CO}_3+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{Na}_2 \mathrm{SO}_4+\mathrm{H}_2 \mathrm{CO}_3

m moles of \mathrm{Na}_2 \mathrm{CO}_3 = m moles of \mathrm{H}_2 \mathrm{SO}_4

= 20 \times 0.1

= 2

m moles of \mathrm{Na}_2 \mathrm{CO}_3 in 300 m L of solution 

\begin{aligned} & =\frac{300}{25} \times 2 \\ \\& =24 \end{aligned}

Weight of \mathrm{Na}_2 \mathrm{CO}_3  =24 \times 106 \times 10^{-3}

=2.544 \; \text{g}

% of  \mathrm{Na}_2 \mathrm{CO}_3=\frac{2.544}{8} \times 100

= 31.8

% of \mathrm{Na}_2 \mathrm{SO}_4=68.2 \%

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