JEE Chemistry 2024 Question with Solution

Given: The salts are $\text{NaBr}$, $\text{NaNO}_3$, $\text{KI}$, and $\text{CaF}_2$. We need the salt that does not evolve colored vapours on heating with concentrated $\text{H}_2\text{SO}_4$.

Find: The molar mass of that salt.

From the solution discussion:

• $\text{NaBr}$ releases $\text{Br}_2$, which gives colored vapours.
• $\text{KI}$ releases $\text{I}_2$, which gives colored vapours.
• $\text{NaNO}_3$ is mentioned as giving $\text{NO}_2$ in the alternate approach, which is also colored.
• $\text{CaF}_2$ does not evolve colored vapours with concentrated $\text{H}_2\text{SO}_4$.

Now calculate the molar mass of $\text{CaF}_2$:

Therefore, the molar mass of the required salt is $78 \, \text{g/mol}$. The correct option is A.

Given: Heating the salts with concentrated $\text{H}_2\text{SO}_4$.

Find: Which salt does not evolve colored vapours, and then its molar mass.

Use elimination first:

Both $\text{Br}_2$ and $\text{I}_2$ are colored vapours, so these are rejected.

The second approach in the solution states:

So $\text{NaNO}_3$ is also not the required salt in the final selection. Hence the remaining salt is $\text{CaF}_2$.

Its molar mass is:

Therefore, the correct option is A.