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Stoichiometric Calculations

The volume in...

Question

The volume (in $m L$ ) of $0.1 M A g N O_{3}$ required to completely precipitate the chloride ions present in $30 m L$ of $0.01 M$ of $[C r (H_{2} O)_{5} C l] C l_{2}$ , as silver chloride is close to:

4 m L

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5 m L

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6 m L

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7 m L

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Solution

The correct option is C $6 m L$
$2 A g N O_{3} n_{1} = 2 + [C r (H_{2} O)_{5} C l n_{2} = 1] C l_{2} \to [C r (H_{2} O)_{5} C l] (N O_{3})_{2} + 2 A g C l$
$\frac{M_{1} V_{1}}{n_{1}} = \frac{M_{2} V_{2}}{n_{2}}$
$\frac{0.1 \times V}{2} = \frac{0.01 \times 30}{1}$
$V = 6 m L$ .

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