The density of 1.48 mass percent calcium hydroxide solution is 1.025 g mL^-1. The volume of 0.1 M HCI solutions required to neutralise 25 mL of this solution is

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90.0 mL

95.5 mL

102.5 mL

85.5 mL

answer is A.

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Detailed Solution

$Mw of Ca {(OH)}_{2} = 74 g Ew of Ca {(OH)}_{2} = \frac{74}{2} = 37 g$

$\begin{array}{l} N_{Ca (OH)_{2}} = \frac{% by weight \times 10 \times d}{{Ew}_{2}} \\ = \frac{1 .48 \times 10 \times 1 .025}{37} = 0 .41 \end{array}$

mEq of Ca(OH)₂ $\equiv$ mEq of HCl

$25 mL \times 0.41 N \equiv 0.1 \times V_{HCl}$

$V_{HCl} = 102.5 mL$

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