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Chemistry · Ionic Equilibrium

JEE Main 2025 — 22 January, Morning Shift — Question 44

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 x , then x 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.]

Answer: 45

Numerical answer — enter this value.

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=\mathrm{Ca}(\mathrm{OH})_{2}= gm equivalent of HCl

n×2=0.1×401000×1\mathrm{n} \times 2=0.1 \times \frac{40}{1000} \times 1

n=2×10−3\mathrm{n}=2 \times 10^{-3}

Volume of CO2=2×10−3×22400=44.8 cm3\mathrm{CO}_{2}=2 \times 10^{-3} \times 22400=44.8 \mathrm{~cm}^{3}

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
22 January, Morning Shift
Subject
Chemistry
Chapter
Ionic Equilibrium
Topic
Solutions with mixture of acids or bases

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