Question

0.50 g sample of impure $CaCO_{3}$ is dissolved in 50 ml of 0.0985 (N) HCl. After the reaction is complete, the excess HCl required 6 ml of 0.105 N NaOH for neutralisation. The percentage purity of $CaCO_{3}$ in the sample is

• A. 42.95
• B. 429.5
• C. 4.295
• D. 21.86

Answer 1

Answer: (A)

Miliequivalents = Normality $\times $ $V_{\left(\right. i n m l \left.\right)}$

Miliequivalents of $HCl$ = Miliequivalents of $NaOH$ $+$ Miliequivalents of $CaCO_{3}$

$0.0985\times 50=0.105\times 6+\frac{W_{C a C O_{3}}}{100}\times 2\times 1000$

$4.925-0.630=W_{C a C O_{3}}\times 20$

$W_{C a C O_{3}}=0.21475g$

$\%CaCO_{3}=\frac{0.21475}{0.50}\times 100=42.95\%$