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The molarity of ${Cl}^{-}$ in an aqueous solution which was (w/v) 2% NaCl, 4% ${CaCl}_{2}$ and 6% ${NH}_{4} Cl$ will be:

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0.342

0.721

2.18

1.12

answer is D.

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Detailed Solution

Molarity = (weight of solute $x$ 1000) / (Molar mass of solute $\times$ volume of solution)

$[\frac{\frac{2}{58.5} + \frac{4}{111} \times 2 + \frac{6}{53.5}}{100}] \times 1000 = 2.18$

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