JEE Chemistry 2025 Question with Solution

Given: Moles of $\mathrm{CO_2}$ = $n$. The entire amount of $\mathrm{CO_2}$ reacted with exactly half the initial amount of $\mathrm{Ca(OH)_2}$.

Find: The volume $x$ of $\mathrm{CO_2}$ in $\text{cm}^3$.

Let the initial moles of $\mathrm{Ca(OH)_2}$ be $2n$. Then the excess unreacted $\mathrm{Ca(OH)_2}$ is $n$.

The gram equivalent of excess $\mathrm{Ca(OH)_2}$ is equal to the gram equivalent of $\mathrm{HCl}$ used in neutralization.

At $273 \, \text{K}$ and $1 \, \text{atm}$, molar volume is $22400 \, \text{cm}^3$ per mole. Therefore,

Therefore, the volume of $\mathrm{CO_2}$ is approximately $45 \, \text{cm}^3$.

Given: Excess $\mathrm{Ca(OH)_2}$ is neutralized by $0.1 \, \text{M}$ $40 \, \text{mL}$ $\mathrm{HCl}$.

Find: Moles of $\mathrm{CO_2}$ and hence its volume.

From the statement, $\mathrm{CO_2}$ reacts with exactly half the initial $\mathrm{Ca(OH)_2}$. So if moles of $\mathrm{CO_2}$ are $n$, then initial $\mathrm{Ca(OH)_2}$ must be $2n$, leaving excess $n$.

Now use neutralization equivalents:

• One mole of $\mathrm{Ca(OH)_2}$ provides $2$ equivalents.
• One mole of $\mathrm{HCl}$ provides $1$ equivalent.

Moles of $\mathrm{HCl}$ used:

So equivalents of $\mathrm{HCl}$ are $4 \times 10^{-3}$, which equals equivalents of excess $\mathrm{Ca(OH)_2}$. Thus,

Hence moles of $\mathrm{CO_2}$ are $2 \times 10^{-3}$. Using molar volume at STP:

Therefore, the nearest integer value of $x$ is 45.