Question

A sample weighing $2.198$ g contains a mixture of $AO$ and $A_{2}S{O}_3$. It takes $0.015$ mol of $K_{2}C{r}_2{O}_7$ to oxidize the sample completely to form $A{O}_4^{-}$ and $C{r}^{3+}$. If $0.0187$ mol of $A{O}_4^{-}$ is formed, what is atomic mass of $A$?

• A. $100$
• B. $120$
• C. $150$
• D. $200$

Answer 1

The correct option is A $100$

Let molar mass of $AO$ and $A_2{O}_3$ be $m$ and $n$ respectively.
$\therefore m=a+16...\left(i\right)$
$n=2+48...\left(ii\right)$
Where, $a$ is atomic mass of $A$
Assume $x$ and $y$ g of $AO$ and $A_2{O}_3$ are present in mixture then
$x+y=2.198...\left(iii\right)$
Since, redox changes are :
$A^{2+}⟶A^{2+}+5e$
$(A^{3+}{)}_2⟶2A^{7+}+8e$
$6e+(C{r}^{6+}{)}_2⟶2C{r}^{3+}$
Also, Meq. of $AO+$Meq. of $A_2{O}_3=$ Meq. of $K_{2}C{r}_2{O}_7$
$\frac{x}{\left(\frac{a+16}{5}\right)}\times 1000+\frac{y}{\left(\frac{2a+48}{8}\right)}\times 1000=0.015\times 6\times 1000$
$\therefore \left(\frac{5x}{a+16}\right)+\left(\frac{8y}{2a+48}\right)=0.09...\left(iv\right)$
Also, mole of $A{O}_4^{-}$ by $AO+$ mole of $A{O}_4^{-}$ by $A_2{O}_3=0.0187$
$\therefore \left(\frac{x}{a+16}\right)+\left(\frac{2y}{2a+48}\right)=\mathrm{0.0187...}\left(v\right)$
($\because$ Mole ratio of $AO:A{O}_4^{-}::1:1$; $A_2{O}_3:A{O}_4^{-}::1:2)$
Solving Eqs.$\left(iii\right),\left(iv\right)and\left(v\right),wegeta=100$