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JEE Advanced levelOne or more correct optionsDifficulty 3 of 3eq218

$1.00 g$ of marble chips (impure $CaCO X_{3}$ ) is dissolved in $40.0 mL$ of $0.500 M$ $HCl$ (excess). The unreacted acid is back-titrated with $0.250 M$ $NaOH$ and needs $22.0 mL$ . Take $CaCO X_{3} = 100$ and molar volume $22.4 L$ at STP. Which statements are correct?

AThe marble is $72.5%$ $CaCO X_{3}$ by mass.
BThe mass of $CaCO X_{3}$ in the marble is $0.725 g$ .
CThe volume of $CO X_{2}$ evolved at STP is $324.8 mL$ .
DThe $HCl$ consumed by the marble is $14.5$ mmol.

Answer: A, B, D

HCl taken

= 20.0

mmol; excess

= 22.0 \times 0.250 = 5.50

mmol; used

= 14.50

mmol.

CaCO X_{3}

= 7.250

mmol

= 0.725 g

, i.e. 72.5\%.

CO X_{2}

= 7.250

mmol

\times 22.4 = 162.4 mL

, not twice this.

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