What mass of $C a C O_{3}$ is required to react completely with 25 ml of 0.75 M $H C l$ ?

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$9.375 \times 10^{- 3} \times 100$

$9.375 \times 10^{- 6} \times 100$

$9.375 \times 10^{- 5} \times 100$

$9.375 \times 10^{- 4} \times 100$

answer is A.

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Detailed Solution

$C a C O_{3} (s) + 2 H C l (a q) \to C a C l_{2} (a q) + H_{2} O (a q) + C O_{2} (g)$
From the above reaction
1 mole $C a C O_{3} = 2 m o l H C l;$ No.of moles of $H C l = \frac{M V}{1000} \times \frac{0.75 \times 25}{1000} = 0.01875$
Number of moles of $C a C O_{3} = \frac{1}{2} \times n o . o f m o l e s o f H C l = \frac{1}{2} \times 0.01871 = 9.375 \times 10^{- 3}$
Mass of $C a C O_{3} =$ number of moles $\times$ molar mass $= 9.375 \times 10^{- 3} \times 100$