0.5 g of a sample of limestone was dissolved in 30ml of N/10 HCl and the solution diluted to 100 ml. 25 ml of this solution required 12.5mL of N/25 NaOH for complete neutralization. Calculate the percentage of $CaC{O}_3$  in the sample.

$CaC{O}_3+2HCl\to CaC{l}_2+H_{2}C{O}_3$

$H_{2}C{O}_3+2NaOH\to N{a}_{2}C{O}_3+2H_{2}O$

To find the % of $CaC{o}_3$ present in the limestone let it be $a\%$

So, in a 0.5 g of sample, $CaC{o}_3$ present $=0.5\times a\%$

$\begin{array}{l}CaC{o}_3=\frac{0.5a}{100}g\\ molesofCaC{o}_3=\frac{0.5a}{100\times 100}=0.5a\times {10}^{-4}mol\end{array}$

Moles of   $H_{2}C{o}_3=molesofCaC{o}_3$

So if 2.5 ml is taken out of 100ml

$\therefore$ moles of  $H_{2}C{o}_3=\frac{molesofCaCo3}{4}=\frac{0.5a\times {10}^{-4}}{4}$

Also, no.of moles of $NaOH$ used = vol normality

$=\frac{12.5ml}{25\times {10}^3}\times \frac{1}{25}$

Also, from the reaction

No. of moles of $H_{2}C{o}_3=\frac{1}{2}\times \left(no.ofmolesofNaOH\right)$

$\begin{array}{l}\frac{0.5a\times {10}^{-4}}{4}=\frac{1}{2}\times \frac{12.5}{25\times {10}^3}\\ a=\frac{4\times {10}^5}{4\times {10}^3\times 5}=20\%\\ \therefore \%gcac{o}_3=20\%\end{array}$