Q78 / 180 · AIIMS 2011 · Chemistry
How much amount of $\mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O}$ is required for liberation of $2.54 \mathrm{~g}$ of $\mathrm{I}_2$ when titrated with $\mathrm{KI}$ ?
- A$2.5 \mathrm{~g}$
- Correct answerB$4.99 \mathrm{~g}$
- C$2.4 \mathrm{~g}$
- D$1.2 \mathrm{~g}$
Solution
$$
\begin{aligned}
& \text { } 2 \mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O}+4 \mathrm{KI} \longrightarrow \\
& \mathrm{Cu}_2 \mathrm{I}_2+2 \mathrm{~K}_2 \mathrm{SO}_4+\mathrm{I}_2+10 \mathrm{H}_2 \mathrm{O} \\
& 254 \mathrm{~g} \\
&
\end{aligned}
$$
Molecular weight of $2 \mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O}$
$$
[2(63.5+32+64)+10(18)] \mathrm{g}=499 \mathrm{~g}
$$
$254 \mathrm{~g}$ of $\mathrm{I}_2$ is liberated by $499 \mathrm{~g} \mathrm{CuSO} \cdot 5 \mathrm{H}_2 \mathrm{O}$
$2.54 \mathrm{~g}$ of $\mathrm{I}_2$ will be liberated by $x$ g CuSO$_4 \cdot 5 \mathrm{H}_2 \mathrm{O}$
$$
x=\frac{499}{254} \times 2.54=4.99 \mathrm{~g}
$$
Hence, the correct answer is option: 'B'.