JEE MAIN - Chemistry (2025 - 22nd January Morning Shift - No. 23)
Some $\mathrm{CO}_2$ gas was kept in a sealed container at a pressure of 1 atm and at 273 K . This entire amount of $\mathrm{CO}_2$ gas was later passed through an aqueous solution of $\mathrm{Ca}(\mathrm{OH})_2$. The excess unreacted $\mathrm{Ca}(\mathrm{OH})_2$ was later neutralized with 0.1 M of 40 mL HCl . If the volume of the sealed container of $\mathrm{CO}_2$ was $x$, then $x$ is ____________ $\mathrm{cm}^3$ (nearest integer).
[Given : The entire amount of $\mathrm{CO}_2(\mathrm{~g})$ reacted with exactly half the initial amount of $\mathrm{Ca}(\mathrm{OH})_2$ present in the aqueous solution.]
Answer
45
Explanation
Let moles of $\mathrm{CO}_2=\mathrm{n}$
moles of $\mathrm{Ca}(\mathrm{OH})_2$
total initially $=2 \mathrm{n}$
excess $\mathrm{Ca}(\mathrm{OH})_2=\mathrm{n}$
gm equivalent of $\mathrm{Ca}(\mathrm{OH})_2=$ gm equivalent of HCl
$$ \begin{aligned} & \mathrm{n} \times 2=0.1 \times \frac{40}{1000} \times 1 \\ & \mathrm{n}=2 \times 10^{-3} \end{aligned} $$
Volume of $\mathrm{CO}_2=2 \times 10^{-3} \times 22400=44.8 \mathrm{~cm}^3$
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