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Faraday's Second Law

A solution co...

Question

A solution containing 0.1 mol of a metal chloride $M C l_{x}$ required 500 ml of 0.8 m $A g N O_{3}$ solution for complete reaction $M C l_{x} + x A g N O_{3} \to x A g C l + M (N O_{3})_{x}$ . Then the value of x is :

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Solution

The correct option is B 4
$n o . m o l e s o f A g N O_{3} = v o l m e * M o l a r i t y = 0.5 * 0.8 = 0.4$
0.1 mol of a metal chloride $M C l_{x}$ required 0.4 mols of $A g N O_{3}$ to complete the reaction.
Then the mole ratio between $M C l_{x} : A g N O_{3} = 1 : x = 1 : 4$
$∴ x = 4$

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