0.50 g sample of impure C a C O 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 C a C O 3 in the sample is

Question

0.50 g sample of impure C a C O 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 C a C O 3 in the sample is, 42.95, 429.5, 4.295, 21.86

MARKS App · Accepted Answer

Miliequivalents = Normality × V ( i n m l ) Miliequivalents of H C l = Miliequivalents of N a O H + Miliequivalents of C a C O 3 0.0985 × 5 0 = 0.105 × 6 + W C a C O 3 100 × 2 × 1000 4.925 - 0.630 = W C a C O 3 × 20 W C a C O 3 = 0.21475 g % C a C O 3 = 0.21475 0.50 × 100 = 42.95 %

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