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11.1g.of ${CaCl}_{2}$ is present in 100 ml of the aqueous solution. The chloride ion concentration is

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0.5M

0.2M

answer is B.

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Detailed Solution

${CaCl}_{2} \to {Ca}^{+ 2} + 2 {Cl}^{-}$

$M_{{Cl}^{-}} = i \times \frac{w}{M . W} \times \frac{1000}{V_{ml}}$

$M_{{Cl}^{-}} = 2 \times \frac{11.1}{111} \times \frac{1000}{100}$

$= 2 M$

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