Question

0.607 g of a silver salt of tribasic organic acid was quantitatively reduced to 0.37 g of pure Ag.What is the mol. wt. of the acid ?

• A. 207
• B. 210
• C. 351
• D. 324

Answer 1

The correct option is B 210

The organic acid is tribasic, hence, its anion will have a charge of $-3$, $Ag$ has a charge of $+1$.

$\therefore$ Molecular formula of the silver salt will be $A{g}_3(A^{(}3-))$

Assuming tribasic acid is $H_{3}A$

$\underset{0.607g}{\underset{}{A{g}_{3}A}}\stackrel{reduced}{⟶}\underset{0.37g}{\underset{}{3Ag}}$

Moles of $Ag=\frac{0.37}{107.8}=3.43\times {10}^{-3}$ $moles$

$1$ $mole$ of salt gives $3$ $moles$ $Ag$

$\therefore x$ $moles$ of salt gives $343\times {10}^{-3}$ $moles$ of $Ag$

$\therefore x=\frac{3.43\times {10}^{-3}\times 1}{3}=1.144\times {10}^{-3}$ $moles$

$Molesofsalt=\frac{Weightofsalt}{Molecularweightofsalt}$

$\therefore$ MOlecular weight of salt $\left(A{g}_{3}A\right)=\frac{0.607}{1.144\times {10}^{-3}}=530.55$ $g/mol$

$\therefore$ Molecular weight of Acid $\left(H_{3}A\right)=$ Molecular weight of salt $-(3\times$ molecular weight of $Ag)+(3\times$ Molecular weight of $H)=530.55-323.4+3$

Molecular weight of acid $=210.15$