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JEE Main 20179 AprilOnline
Exam
JEE Main
Year
2017
Date
9 April
Mode
Online
Subject
Chemistry
Chapter
Solutions

JEE Main 2017 · 9 April · Online · Chemistry · Solutions

A solution is prepared by mixing 8 . 5 g of CH _ 2 Cl _ 2 and...

A solution is prepared by mixing 8.58 . 5 g of CH2Cl2and11.95gofCHCl3{CH}_{2}{Cl}_{2} and 11.95\text{g} of {CHCl}_{3} . If vapour pressure of CH2Cl2andCHCl3{CH}_{2}{Cl}_{2} and {CHCl}_{3} at 298298K are 415and200415 and 200 mm Hg respectively, the mole fraction of CHCl3{CHCl}_{3} in vapour form is: (MolarmassofCl=35.5gmol−1)\left(\text{M}\text{o}\text{l}\text{a}\text{r}\text{m}\text{a}\text{s}\text{s}\text{o}\text{f}Cl = 35.5\text{g}\text{m}\text{o}{\text{l}}^{- 1}\right)

Options

  1. A0.1620 . 162
  2. B0.6750 . 675
  3. C0.3250 . 325
  4. D0.4860 . 486

Correct answer

C. 0.3250 . 325

Step-by-step solution

mole of CH2Cl2in liquid phase=8.585=0.1{\text{mole of CH}}_{2}\text{C}{\text{l}}_{2}\text{in liquid phase} = \dfrac{8.5}{85} = 0.1

mole of CHCl3in liquid phase=11.95119.5=0.1{\text{mole of CHCl}}_{\text{3}}\text{in liquid phase} = \dfrac{11.95}{119.5} = 0.1

mole fraction of CH2Cl2in liquid phase=0.10.2=12{\text{mole fraction of CH}}_{\text{2}}{\text{Cl}}_{\text{2}}\text{in liquid phase} = \dfrac{0.1}{0.2} = \dfrac{1}{2}

mole fraction of CHCl3in liquid phase=0.10.2=12{\text{mole fraction of CHCl}}_{\text{3}}\text{in liquid phase} = \dfrac{0.1}{0.2} = \dfrac{1}{2}

PT=XCH2Cl2×vapour pressureCH2Cl2+XCHCl3×vapour pressureCHCl3{\text{P}}_{\text{T}} = {\text{X}}_{\text{C}{\text{H}}_{2}\text{C}{\text{l}}_{2}} \times {\text{vapour pressure}}_{\text{C}{\text{H}}_{2}\text{C}{\text{l}}_{2}} + {\text{X}}_{\text{C}\text{H}\text{C}{\text{l}}_{3}} \times {\text{vapour pressure}}_{\text{C}\text{H}\text{C}{\text{l}}_{3}}

=415×12+200×12=307.5= 415 \times \dfrac{1}{2} + 200 \times \dfrac{1}{2} = 307.5

PT(XA)VP=(XA)LP×Vapour pressure of CHCl3{\text{P}}_{\text{T}} {( {\text{X}}_{\text{A}} )}_{\text{V}\text{P}} = {( {\text{X}}_{\text{A}} )}_{\text{L}\text{P}} \times {\text{Vapour pressure of CHCl}}_{3}

307.5×(XCHCl3)VP=200×12307.5 \times {( {\text{X}}_{\text{C}\text{H}\text{C}{\text{l}}_{3}} )}_{\text{V}\text{P}} = 200 \times \dfrac{1}{2}

XCHCl3=100307.5=0.325{\text{X}}_{\text{C}\text{H}\text{C}{\text{l}}_{3}} = \dfrac{100}{307.5} = 0.325

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