Question

A $2.18g$ sample contains a mixture of $XO$ and $X_2{O}_3.$ It reacts with 0.015 moles of $K_{2}C{r}_2{O}_7$ to oxidize the sample completely to form $X{O}_4^{-}$ and $C{r}^{3+}$. If 0.0187 moles of $X{O}_4^{-}$ are formed, what is the atomic mass of X?

• A. $99.08$
• B. $9.08$
• C. $54.0$
• D. $32.0$

Answer 1

The correct option is A $99.08$

$XO+K_{2}C{r}_2{O}_7\to C{r}^{3+}+X{O}_4^{-}$
$X_2{O}_3+K_{2}C{r}_2{O}_7\to C{r}^{3+}+X{O}_4^{-}$
Let, weight of XO in the mixture be 'x' g.
$\text{Equivalents of}{K}_{2}C{r}_2{O}_7\text{consumed by the mixture}=0.015\times 6$
$\text{Equivalents of}XO=\frac{x}{M+16}\times 5$
$\text{Equivalents of}{X}_2{O}_3=\frac{2.18-x}{2M+48}\times 8$

$\text{Equivalents of}XO+\text{Equivalents of}{X}_2{O}_3=\text{Equivalents of}{K}_{2}C{r}_2{O}_7$
$\therefore \frac{x}{M+16}\times 5+\frac{2.18-x}{2M+48}\times 8=0.015\times 6...\left(1\right)$
Since 1 mole of $XO$ gives 1 mole $X{O}_4^{-}$ and 1 mole of $X_2{O}_3$ gives 2 moles of $X{O}_4^{-},$
$\therefore \frac{x}{M+16}+\frac{2\times (2.18-x)}{2M+48}=0.0187...\left(2\right)$
From Equation (1) and (2), we get
$x=1.74gM=99.08g/mol$