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JEE Main 202126 FebruaryMorning Shift
Exam
JEE Main
Year
2021
Date
26 February
Shift
Morning Shift
Subject
Chemistry
Chapter
Solutions

JEE Main 2021 · 26 February · Morning Shift · Chemistry · Solutions

224mL of SO _ 2 ( g ) at 298K and 1atm is passed through 100mL of 0...

224mL224mL of SO2(g){SO}_{2 ( g )} at 298K298K and 1atm1atm is passed through 100mL100mL of 0.1MNaOH0 . 1MNaOH solution. The non-volatile solute produced is dissolved in 36g36g of water. The lowering of vapour pressure of solution (assuming the solution is dilute) (P°(H2O)=24mmP{°}_{\left({H}_{2}O\right)} = 24mm of HgHg) is x×10−2mmx \times {10}^{- 2}mm of HgHg the value of xx is (Integer answer)

Correct answer

0

Step-by-step solution

Sol (1)\left(1\right)

{SO2+2NaOH→Na2SO3+H2O2240.0821×298=9.2mmol10mmol5mmol(L.R.)(i=3)\begin{cases}{SO}_{2} & + & 2NaOH & \to & {Na}_{2}{SO}_{3} & + & {H}_{2}O \\ \dfrac{224}{0 . 0821 \times 298} = 9 . 2mmol & & 10mmol & & 5mmol & & \\ & & ( L . R . ) & & ( i = 3 ) & & \end{cases}

P0−Ps=iXsoluteP0{P}^{0} - {P}^{s} = {iX}_{solute}{P}^{0}

=24×3×5×10−32= 24 \times \dfrac{3 \times 5 \times {10}^{- 3}}{2}

=0.18=18×10−3= 0 . 18 = 18 \times {10}^{- 3} mm of Hg

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