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Salt of Strong Acid and Strong Bases

The molarity ...

Question

The molarity of $C l^{-}$ ion in a 100 mL aqueous solution which has 2 % $(\frac{w}{v}) N a C l$ , 4 % $(\frac{w}{v}) C a C l_{2}$ and 6 % $(\frac{w}{v}) N H_{4} C l$ will be:

0.342 M

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0.721 M

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1.121 M

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2.181 M

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Solution

The correct option is D 2.181 M
100 mL solution contains 2 g $N a C l$ , 4 g $C a C l_{2}$ and 6 g $N H_{4} C l$
Moles of $N a C l = \frac{2}{58.5} = 0.034$
$∴ Moles of C l^{-} from N a C l = 0.034 mol . . . . . . . (i)$
$Moles of C a C l_{2} = \frac{4}{111} = 0.036 mol$
$∴ Moles of C l^{-} from C a C l_{2} = 2 \times 0.036 = 0.072 mol . . . . . . . (i i)$
$Moles of N H_{4} C l = \frac{6}{53.5} = 0.112 mol$
$∴ Moles of C l^{-} from N H_{4} C l = 0.112 mol . . . . . . . (i i i)$
$Molarity = \frac{Moles of C l^{-}}{Volume in L} = \frac{0.034 + 0.072 + 0.112}{0.1}$

$Molarity = 2.18 M$

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Similar questions

The molarity of Cl^- in an aqueous solution which has w/v 2% NaCl, 4% CaCl₂and 6% NH₄Cl will be

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b) 0.721

c) 1.12

d) 2.18

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