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Stoichiometric Calculations

10 g CaCO3 ...

Question

$10$ g $C a C O_{3}$ was dissolved in $250$ mL of $1 M H C l$ and the solution was boiled. The amount of volume (in ml) of $2 M K O H$ that would be required to reach the equivalence point after boiling is (assume no change in volume during boiling) :

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Solution

Meq. of $C a C O_{3}$ dissolved = $\frac{10}{\frac{100}{2}} \times 1000 = 200$
Meq. of $H C l$ added = $250 \times 1 = 250$
$∴$ Meq. of $H C l$ after reaction with $C a C O_{3} = 250 - 200 = 50$
Now, meq. of $H C l$ left $=$ meq. of $K O H$ used
$⟹ 50 = 2 \times V$ $⟹ V = 25$ mL
Hence, volume required is $25$ mL.

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