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JEE Main 202522 JanuaryMorning Shift
Exam
JEE Main
Year
2025
Date
22 January
Shift
Morning Shift
Subject
Chemistry
Chapter
Redox Reactions

JEE Main 2025 · 22 January · Morning Shift · Chemistry · Redox Reactions

Some CO_2 gas was kept in a sealed container at a pressure of 1 atm...

Some CO2\mathrm{CO}_2 gas was kept in a sealed container at a pressure of 1 atm and at 273 K . This entire amount of CO2\mathrm{CO}_2 gas was later passed through an aqueous solution of Ca(OH)2\mathrm{Ca}(\mathrm{OH})_2. The excess unreacted Ca(OH)2\mathrm{Ca}(\mathrm{OH})_2 was later neutralized with 0.1 M of 40 mL HCl . If the volume of the sealed container of CO2\mathrm{CO}_2 was xx, then xx is ________ cm3\mathrm{cm}^3 (nearest integer). [Given : The entire amount of CO2( g)\mathrm{CO}_2(\mathrm{~g}) reacted with exactly half the initial amount of Ca(OH)2\mathrm{Ca}(\mathrm{OH})_2 present in the aqueous solution.]

Correct answer

0

Step-by-step solution

Let moles of CO2=n\mathrm{CO}_2=\mathrm{n} moles of Ca(OH)2\mathrm{Ca}(\mathrm{OH})_2 total initially =2n=2 \mathrm{n} excess Ca(OH)2=n\mathrm{Ca}(\mathrm{OH})_2=\mathrm{n} gm equivalent of Ca(OH)2=gm\mathrm{Ca}(\mathrm{OH})_2=\mathrm{gm} equivalent of HCl n×2=0.1×401000×1n=2×10−3\begin{aligned} & \mathrm{n} \times 2=0.1 \times \frac{40}{1000} \times 1 \\ & \mathrm{n}=2 \times 10^{-3} \end{aligned} Volume of CO2=2×10−3×22400=44.8 cm3\mathrm{CO}_2=2 \times 10^{-3} \times 22400=44.8 \mathrm{~cm}^3
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