The number of $C l^{-}$ and $C a^{+ 2}$ ions in 222g. of $C a C l_{2}$ are

4N, 2N

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2N, 4N

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1N, 2N

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2N, 1N

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Solution

The correct option is A
4N, 2N

No. of molecules = $\frac{222}{111} \times N$ = $2 N$

$C a C l_{2} \to C a^{+ 2} + 2 C l^{-}$

No. of $C l^{-}$ ions = $2 \times 2 N$ = $4 N$

No. of $C a^{+ 2}$ ions = $2 N$

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