JEE Main 2020 — Chemistry Physical Chemistry

2020mcqmedium

Henry's constant (in kbar) for four gases $\alpha ,\beta ,\gamma$ and $\delta$ in water at $298K$ is given below :

$\begin{matrix} & \alpha & \beta & \gamma & \delta \\ {K}_{H} & 50 & 2 & 2\times {10}^{-5} & 0.5\end{matrix}$

(density of water $={10}^{3}k{\mathrm{gm}}^{-3}$ at $298K$ ) This table implies that :

Official previous-year question

Held on 3 Sept 2020 · Verified 6 Jul 2026.

Options

A
$\alpha$ has the highest solubility in water at a given pressure
B
solubility of $\gamma$ at $308K$ is lower than at $298K$
C
The pressure of a $55.5$ molal solution of $\delta$ is $250$ bar
D
The pressure of $55.5$ molal solution of $\gamma$ is $1\mathrm{bar}$ .

Did you get this right?

Solution

(1) From Henery's law $P={k}_{H}(X)$

Higher the value of ${k}_{H}$ smaller will be solubility so $\gamma$ is more soluble.

(2) Though solubility of gas will decrease with increase in temperature but this conclusion can not be drawn from the given table.

(3)For

$\delta \Rightarrow {P}_{\delta }=({K}_{H\delta }).(X)\delta$

$=0.5\times {10}^{3}\times [\frac{55.55}{55.55+\frac{1000}{18}}]=250\mathrm{bar}.$

(4)For $\gamma$

${(p)}_{\gamma }={({K}_{H})}_{\gamma }\times {(x)}_{\gamma }$

$=2\times {10}^{-2}[\frac{55.5}{55.5+\frac{1000}{18}}]={10}^{-2}\mathrm{bar}$

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Options

$\alpha$ has the highest solubility in water at a given pressure

solubility of $\gamma$ at $308K$ is lower than at $298K$

The pressure of a $55.5$ molal solution of $\delta$ is $250$ bar

The pressure of $55.5$ molal solution of $\gamma$ is $1\mathrm{bar}$ .

Solution

(1) From Henery's law $P={k}_{H}(X)$

Higher the value of ${k}_{H}$ smaller will be solubility so $\gamma$ is more soluble.

(2) Though solubility of gas will decrease with increase in temperature but this conclusion can not be drawn from the given table.

(3)For

$\delta \Rightarrow {P}_{\delta }=({K}_{H\delta }).(X)\delta$

$=0.5\times {10}^{3}\times [\frac{55.55}{55.55+\frac{1000}{18}}]=250\mathrm{bar}.$

(4)For $\gamma$

${(p)}_{\gamma }={({K}_{H})}_{\gamma }\times {(x)}_{\gamma }$

$=2\times {10}^{-2}[\frac{55.5}{55.5+\frac{1000}{18}}]={10}^{-2}\mathrm{bar}$

JEE Main 2020 — Chemistry Physical Chemistry

Options

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Solution

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JEE Main Physical Chemistry in other years

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JEE Main 2020 — Chemistry Physical Chemistry

Options

Did you get this right?

Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

Organic Chemistry

Inorganic Chemistry

Keep practicing Physical Chemistry