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JEE Main levelSingle correct optionDifficulty 2 of 3eq132

25.0 mL of saturated lime water ( $Ca (OH) X_{2}$ ) needs 18.0 mL of 0.0500 M HCl for neutralisation. Using molar mass 74 g/mol, the solubility of $Ca (OH) X_{2}$ in g/L is:

A $2.66 g/L$
B $1.33 g/L$
C $0.67 g/L$
D $0.0180 g/L$

Answer: B

HCl = 0.0500

\times

18.0 = 0.900 mmol

H X^{+}

, and each

Ca (OH) X_{2}

gives 2

OH X^{-}

, so

Ca (OH) X_{2}

= 0.450 mmol in 25.0 mL, i.e. 0.0180 M. Solubility = 0.0180

\times

74 = 1.33 g/L.

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