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  3. Chemistry
  4. 224 mL of SO 2( g ) at 298 K and 1 atm is passed through 100 mL of 0.1 M NaOH solution. The non-volatile solute produced is dissolved in 36 g of water. The lowering of vapour pressure of solution (assuming the solution is dilute) ( P ( H 2 O )=24 mm . of . Hg ) is x × 10-2 mm of Hg the value of x is .

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

JEE MainJEE Main 2021Solutions

Solution:

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$= 9.2\, m \,mol$

$ P ^{s} = P ^{0} \cdot X _{\text {solvent }} $

$ =24 \times \frac{2}{\left(2+15 \times 10^{-3}\right)} $

$ =23.82$

$\Delta P =0.18$ torr

$=18 \times 10^{-2}$ torr

image

$\Delta P = P ^{0} \cdot X _{\text {solute }} $

$=24 \times \frac{(1.6+18.4)}{2020}$

$=0.2376=23.76 \times 10^{-2} $