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Molarity

A 100 mL so...

Question

A $100 m L$ solution containing $A g N O_{3}$ was treated with excess $N a C l$ to completely precipitate the silver as $A g C l$ . If $5.7 g A g C l$ was obtained, what was the concentration of $A g^{+}$ in the original solution? (At. mass of $A g = 108 g / m o l$ )

0.03 M

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0.05 M

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0.12 M

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0.40 M

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Solution

The correct option is C $0.40 M$
$A g N O_{3} + N a C l \to A g C l ↓ + N a N O_{3}$
$1 m o l e$ of $A g^{+}$ ions produce $1 m o l e$ of $A g C l$ precipitate.
Molar mass of $A g C l = 108 + 35.5 = 143.5 g$
So, $1 m o l e$ $A g^{+} \to$ $143.5 g A g C l$
$(x)$ $\to$ $5.7 g A g C l$
$\Rightarrow x = \frac{5.7}{143.5} = 0.04 m o l e s$ of $A g^{+}$
Molarity of $A g^{+} = \frac{No. of moles}{V o l u m e} = \frac{0.04}{0.1 L} = 0.4 M$

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