The normality of $1.11 g$ of $C a C l_{2}$ dissolved in $250 m L$ solution is .

0.1 N

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0.25 N

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0.08 N

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0.3 N

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Solution

The correct option is C $0.08 N$
Normality = $\frac{G r a m e q u i v a l e n t s o f s o l u t e}{V o l u m e o f s o l u t i o n (i n l i t r e)}$

Equivalent mass $= \frac{M o l e c u l a r M a s s}{C h a r g e o n i o n \times n u m b e r o f i o n s}$
[ $C a \to {C a}^{2 +} + 2 e^{-}$ ]
[ $C l_{2} + 2 e^{-} \to 2 C l^{-}$ ]
Either way, the denominator would be 2, please check this carefully!

= $\frac{111}{2}$ = 55.5
No of gram equivalents = $\frac{1.11}{55.5}$
$\Rightarrow$ Normality = $\frac{\frac{1.11}{55.5}}{\frac{250}{1000}} = 0.08 N$

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