$0.304 g$ of a silver salt of a dibasic acid left $0.216 g$ of silver on ignition. Calculate its molecular weight.

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Solution

If the molecular formula is $A g_{2} A (A^{2 -}$ is dibasic away)

$A g_{2} A + 2 e^{-} \to 2 A g .$
$0.304 g$ $0.216 g$

$2 \times m o l e s o f A g_{2} A =$ moles of $A g .$

$\frac{2 \times 0.304 g}{M o l w t o f A g_{2} A} = \frac{0.216}{108} = 0.002$

Mol. wt of $A g_{2} A = \frac{0.304}{0.001} = 304 g$

$2 \times 108 + w t o f A^{2 -} = 304 g$

Wt. of $A^{2 -} = 304 - 216 = 88 g$

Also molecular formula of acid $= H_{2} A$

$∴$ Molecular weight of acid $= 2 + 88 = 90 g$

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