$0.900 g$ of powdered eggshell (assume only $CaCO X_{3}$ reacts) is treated with $50.0 mL$ of $0.250 M$ $HCl$ and the mixture is diluted to exactly $250.0 mL$ . A $25.0 mL$ portion of this solution needs $3.4 mL$ of $0.100 M$ $NaOH$ . Taking $CaCO X_{3} = 100$ , find the percentage of $CaCO X_{3}$ in the eggshell (one decimal).

Numerical answer type, answer in %, tolerance 0.1.

Answer: 50.6 %

Excess HCl in the aliquot

= 3.4 \times 0.100 = 0.340

mmol, so in the whole

250.0 mL

it is

3.40

mmol. HCl used

= 12.50 - 3.40 = 9.10

mmol.

CaCO X_{3}

= 9.10/2 = 4.550

mmol

= 455.0

mg. Percentage

= 455.0/900 \times 100 = 50.56%

, i.e.

50.6%