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JEE Main 20199 AprilMorning Shift
Exam
JEE Main
Year
2019
Date
9 April
Shift
Morning Shift
Subject
Chemistry
Chapter
Solutions

JEE Main 2019 · 9 April · Morning Shift · Chemistry · Solutions

The osmotic pressure of a dilute solution of an ionic compound XY...

The osmotic pressure of a dilute solution of an ionic compound XYXY in water is four times that of a solution 0.01MBaCl20.01MBaC{l}_{2} in water. Assuming complete dissociation of the given ionic compounds in water, the concentration of XYXY ( inmolL−1inmol{L}^{- 1} ) in solution is

Options

  1. A4×10−24 \times {10}^{- 2}
  2. B16×10−416 \times {10}^{- 4}
  3. C4×10−44 \times {10}^{- 4}
  4. D6×10−26 \times {10}^{- 2}

Correct answer

D. 6×10−26 \times {10}^{- 2}

Step-by-step solution

Osmotic pressure π=i×CRT\pi = i \times CRT

i=i =Van't hoff factor

According to question,

for XY i=2,BaCl2i = 2, BaC{l}_{2} i=3i = 3

πXY=4×πBaCl2{\pi}_{XY} = 4 \times {{\pi}_{BaCl}}_{2}

2×C×RT=4×3×0.01×RT2 \times C \times RT = 4 \times 3 \times 0.01 \times RT

C=6×0.01=0.06MC = 6 \times 0.01 = 0.06M

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