If 0.228 , g of silver salt of a dibasic acid | vedclass

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(C) Step $1$: Calculate the equivalent mass of the silver salt $(E_{salt})$.Using the relation: $\frac{	ext{Mass of salt}}{	ext{Mass of Ag}} = \frac

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$90$

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(C) Step $1$: Calculate the equivalent mass of the silver salt $(E_{salt})$.Using the relation: $\frac{	ext{Mass of salt}}{	ext{Mass of Ag}} = \frac{	ext{Eq. mass of salt}}{	ext{Eq. mass of Ag}}$$\frac{0.228}{0.162} = \frac{E_{salt}}{108}$$E_{salt} = \frac{0.228}{0.162} 	imes 108 = 152 \, g/eq$.Step $2$: Calculate the equivalent mass of the acid $(E_{acid})$.The silver salt of a dibasic acid $(H_2A)$ is $Ag_2A$. The equivalent mass of the salt is $E_{acid} + E_{Ag} - 1$ (since $2$ $H^+$ are replaced by $2$ $Ag^+$).$E_{salt} = E_{acid} + 108 - 1 = E_{acid} + 107$.$152 = E_{acid} + 107 \implies E_{acid} = 45 \, g/eq$.Step $3$: Calculate the molecular mass of the acid.$	ext{Molecular mass} = 	ext{Equivalent mass} 	imes 	ext{Basicity} = 45 	imes 2 = 90$.

If $0.228 \, g$ of silver salt of a dibasic acid gave a residue of $0.162 \, g$ of silver on ignition,then the molecular weight of the acid is:

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