224 mL of SO_2(g) at 298K and… - JEE Main 2021 Chemistry PYQ

JEE Main 2021 — Chemistry Physical Chemistry

2021integerhard

$224\mathrm{mL}$ of ${\mathrm{SO}}_{2(g)}$ at $298K$ and $1\mathrm{atm}$ is passed through $100\mathrm{mL}$ of $0.1M\mathrm{NaOH}$ solution. The non-volatile solute produced is dissolved in $36g$ of water. The lowering of vapour pressure of solution (assuming the solution is dilute) ( $P{^{\circ}}_{({H}_{2}O)}=24\mathrm{mm}$ of $\mathrm{Hg}$ ) is $x\times {10}^{-2}\mathrm{mm}$ of $\mathrm{Hg}$ the value of $x$ is (Integer answer)

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Held on 26 Feb 2021 · Verified 6 Jul 2026.

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Solution

Sol $(1)$

$\begin{matrix}{\mathrm{SO}}_{2} & + & 2\mathrm{NaOH} & \rightarrow & {\mathrm{Na}}_{2}{\mathrm{SO}}_{3} & + & {H}_{2}O \\ \frac{224}{0.0821\times 298}=9.2m\mathrm{mol} & & 10\mathrm{mmol} & & 5\mathrm{mmol} & & \\ & & (L.R.) & & (i=3) & & \end{matrix}$

${P}^{0}-{P}^{s}={\mathrm{iX}}_{\mathrm{solute}}{P}^{0}$

$=24\times \frac{3\times 5\times {10}^{-3}}{2}$

$=0.18=18\times {10}^{-3}$ mm of Hg

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Solution

Sol $(1)$

${P}^{0}-{P}^{s}={\mathrm{iX}}_{\mathrm{solute}}{P}^{0}$

$=24\times \frac{3\times 5\times {10}^{-3}}{2}$

$=0.18=18\times {10}^{-3}$ mm of Hg

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Solution

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Solution

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JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

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Inorganic Chemistry

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