CGPET 2018 Solved Paper

(a) Among the given options: $\because \mathrm{FeSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O}$ is of light green colour. $\therefore \mathrm{FeSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O}$ is $(A)$ (b) $3 \mathrm{FeSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O} + \text{ dil. } 4 \mathrm{HNO}_{3} \rightarrow$ $\mathrm{Fe}(\mathrm{NO}_{3})_{3} + \mathrm{Fe}_{2}(\mathrm{SO}_{4})_{3} + \mathrm{NO} + 23 \mathrm{H}_{2} \mathrm{O}$ The yellow solution is due to $\mathrm{Fe}(\mathrm{NO}_{3})_{3}$. Note : In options, $\mathrm{FeCl}_{3}$ was given which was an error and thus, has replaced by $\mathrm{Fe}(\mathrm{NO}_{3})_{3}$. (c) $\mathrm{FeSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O}$ gives white ppt. with $\mathrm{BaCl}_{2}$. $\mathrm{FeSO}_{4} \cdot 7 \mathrm{KH}_{2} \mathrm{O} + \mathrm{BaCl}_{2} \rightarrow$ $\mathrm{FeCl}_{2} + \mathrm{BaSO}_{4} + 7 \mathrm{KH}_{2} \mathrm{O}$ $\therefore (c)$ is $(\mathrm{BaSO}_{4})$ (d) When $\mathrm{FeSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O}$ is heated, it looses its water and give brown mass of $\mathrm{Fe}_{2} \mathrm{O}_{3}, \mathrm{SO}_{2}(g)$ and $\mathrm{SO}_{3}(g)$ as follows. $2 \mathrm{FeSO}_{4} \xrightarrow{\Delta} \mathrm{Fe}_{2} \mathrm{O}_{3}(s)$ $+ \mathrm{SO}_{2}(g) + \mathrm{SO}_{3}(g)$ $\therefore (d)$ is $(\mathrm{Fe}_{2} \mathrm{O}_{3})$ $\therefore (b)$ is the correct option.