0.3 mol of NaCl and 0.3 mol of $C a C l_{2}$ were dissolved in water to produce a 800 ml solution. Calculate the molarity of the Cl ion.

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Solution

$N a C l \to N a^{+} + C l^{-}$
$0.3$ Mole f the $N a C l$ gives $0.3$ Mole of $C l$ ion $.$
$C a C l_{2} \to C a_{2}^{+} + 2 C l^{-}$
$0.3$ Mole of $C a C l_{2}$ gives $0.6$ Mole of $C l$ ion $.$
$∴$ Total mole of $C l^{-}$ ion $= 0.3 + 0.6 = 0.9$
$∴ M = \frac{T o t a l N u m b e r o f M o l e s}{M o l a r i t y o f t h e s o l u t i o n}$
$= \frac{0.9}{0.8} = 1.125 M$

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