Question

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

Answer 1

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

$A{g}_{2}A+2e^{-}\to 2Ag.$

$0.304g$ $0.216g$

$2\times molesofA{g}_{2}A=$ moles of $Ag.$

$\frac{2\times 0.304g}{MolwtofA{g}_{2}A}=\frac{0.216}{108}=0.002$

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

$2\times 108+wtof{A}^{2-}=304g$

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

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

$\therefore$ Molecular weight of acid $=2+88=90g$