Consider a solution of CO2 (g) dissolved in water in a closed con… — Solution | JEE Chemistry

MCQEasyJEE 2026Raoult's Law & Vapour Pressure

Consider a solution of $\mathrm{CO_2}$$$$\text{(g)}$ dissolved in water in a closed container. Which one of the following plots correctly represents variation of $\log$ (partial pressure of $\mathrm{CO_2}$ in vapour phase above water) [y-axis] with $\log$ (mole fraction of $\mathrm{CO_2}$ in water) [x-axis] at $25^\circ \text{C}$ ?

$Four graph options are missing for a log partial pressure versus log mole fraction plot of carbon dioxide dissolved in water at 25 degree Celsius.$

A
Option A
B
Option B
C
Option C
D
Option D

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: A solution of $\mathrm{CO_2}$ dissolved in water in a closed container at $25^\circ \text{C}$ . We need the relation between partial pressure in vapour phase and mole fraction in water.

Find: Which plot of $\log P$ versus $\log x$ is correct.

By Henry's law,

P = k_H x

where $P$ is the partial pressure of the gas and $x$ is its mole fraction in solution.

Taking logarithm on both sides,

\log P = \log k_H + \log x

This is of the form

y = x + c

So the graph of $\log P$ versus $\log x$ is a straight line with slope $1$ and a non-zero intercept.

Therefore, the correct option is A.

Common mistakes

Assuming the relation between $P$ and $x$ is non-linear. This is incorrect because Henry's law gives direct proportionality at constant temperature. Use $P = k_H x$ before taking logarithms.
Forgetting that taking logarithm of $P = k_H x$ produces an intercept term. This is wrong because $\log k_H$ shifts the line vertically. The graph is not forced to pass through the origin on the log-log plot.
Confusing the slope of the log-log graph with the value of Henry's constant. This is incorrect because the slope comes from the exponent of $x$ and is $1$ here, while $k_H$ only affects the intercept.

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Step-by-step solution

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