Question

$0.015$ mol of $K_{2}C{r}_2{O}_7$ oxidises $2.18$ g of a mixture of $XO$ and $X_2{O}_3$ into ${X{O}_4}^{-}$ in acidic medium. If $0.0187$ mol of ${X{O}_4}^{-}$ is formed, then calculate the atomic weight of $X$.
$6XO+5{C{r}_2{O}_7}^{2-}+34H^{+}\to 6{X{O}_4}^{-}+10C{r}^{3+}17H_{2}O$
$3X_2{O}_3+4{C{r}_2{O}_7}^{2-}+26H^{+}\to 6{X{O}_4}^{-}+8C{r}^{3+}13H_{2}O$

Answer 1

Let $M$ be the atomic weight of $X$.
The molar mass of $XO$ is $(16+M)g/mol$
The molar mass of $X_2{O}_3$ is $(20M+48)g/mol$
Let the mixture contains $w$ g of $XO$ and $(2.18-w)$ g of $X_2{O}_3$.
$6$ moles of $XO$ will require $5$ moles of $K_{2}C{r}_2{O}_7$.
Hence, $\frac{w}{16+M}$ moles of $XO$ will require $\frac{w}{16+M}\times \frac{5}{6}$ moles of $K_{2}C{r}_2{O}_7$.
$3$ moles of $X_2{O}_3-$ will require $4$ moles of $K_{2}C{r}_2{O}_7$.
Hence, $\frac{2.18-w}{2M+48}$ moles of $X_2{O}_3$ will require $\frac{2.18-w}{2M+48}\times \frac{4}{3}$ moles of $K_{2}C{r}_2{O}_7$.
Total number of moles of $K_{2}C{r}_2{O}_7$ required $=\frac{w}{16+M}\times \frac{5}{6}+\frac{2.18-w}{2M+48}\times \frac{4}{3}=0.015mol$ ......(1)
$6$ moles of $XO$ will form $6$ moles of ${X{O}_4}^{-}$.
$\frac{w}{16+M}$ moles of $XO$ will form $\frac{w}{16+M}$ moles of ${X{O}_4}^{-}$.
$3$ moles of $X_2{O}_3$ will form $6$ moles of ${X{O}_4}^{-}$.
$\frac{2.18-w}{2M+48}$ moles of $X_2{O}_3$ will form $2\times \frac{2.18-w}{2M+48}$ moles of ${X{O}_4}^{-}$.
Total moles of ${X{O}_4}^{-}$ formed is $\frac{w}{16+M}+2\times \frac{2.18-w}{2M+48}$ = $0.0187mol$ ......(2)
From (1) and (2), $w=1.7372g$ and $M=9g/mol$.