Question

$0.607g$ of the silver salt of a tribasic acid on combustion deposited $0.37g$ of pure silver. Calculate the molecular weight of the acid.
Enter the nearest integer value.

Answer 1

Formula of silver salt $=A{g}_{3}A$(A is trivalent anion)

$A{g}_{3}A\to 3Ag$

$0.607g$ $0.37g$

$3\times$moles of$A{g}_{3}A=$ moles of $Ag$

$\frac{3\times 0.607}{Mol.wt.A{g}_{3}A}=\frac{0.37}{108}$

Mol. wt $A{g}_{3}A=\frac{3\times 0.607\times 108}{0.37}=531.54$

Wt. of $A^{3-}=531.54-3\times 108=531.54-3204=207.54$

Mol. formula of acid $=H_{3}A$

Mol. wt $=3\times 1+207.54=210.54g\approx 211g$