When heated cautiously in a crucible, $0.759 \mathrm{g}$ of a silver salt of a dibasic acid gave $0.463 \mathrm{g}$ of metallic silver. Calculate the molar mass of the acid. 

Equivalent weight of silver salt is calculated as:

$\begin{array}{ccc}\mathrm{Eqquivalent} \mathrm{weight} \mathrm{of} \mathrm{silver} \mathrm{salt}& =& \frac{\mathrm{Weight} \mathrm{of} \mathrm{silver} \mathrm{salt}}{\mathrm{Weight} \mathrm{of} \mathrm{silver} \mathrm{as} \mathrm{residue}}\times \mathrm{Equivalent} \mathrm{weight} \mathrm{of} \mathrm{Ag}\\ & =& \frac{0.759}{0.463}\times 108\\ & =& 177\end{array}$

As silver salt of dibasic, as a result the n-factor of silver salt is 2.

$\begin{array}{ccc}\mathrm{Equivalent} \mathrm{weight} \mathrm{of} \mathrm{acid}& =& \mathrm{Equivalent} \mathrm{weight} \mathrm{of} \mathrm{silver} \mathrm{salt}-\mathrm{equivalent} \mathrm{weight} \mathrm{of} \mathrm{Ag}+\mathrm{equivalent} \mathrm{weight} \mathrm{of} \mathrm{H}\\ & =& 177-108+1\\ & =& 70\end{array}$

Relation between equivalent weight and molecular weight of acid.

$\begin{array}{ccc}\mathrm{Molecular} \mathrm{weight}& =& \mathrm{Equivalent} \mathrm{weight}\times \mathrm{n}-\mathrm{factor}\\ & =& 70\times 2\\ & =& 140\end{array}$