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NEET 2021
Exam
NEET
Year
2021
Subject
Chemistry
Chapter
Solutions

NEET 2021 · Chemistry · Solutions

The following solutions were prepared by dissolving 10g of glucose...

The following solutions were prepared by dissolving 10g10g of glucose \left({C}_{6}{H}_{12}{O}_{6}\right) in 250mLofwater(P1),10gof water \left({P}_{1}\right), 10g of urea \left({CH}_{4}{N}_{2}O\right) in 250mLofwater(P2)and10gof water \left({P}_{2}\right) and 10g of sucrose \left({C}_{12}{H}_{22}{O}_{11}\right) in 250mLofwater(P3).of water \left({P}_{3}\right) . The right option for the decreasing order of osmotic pressure of these solutions is:

Options

  1. AP2>P1>P3{P}_{2} > {P}_{1} > {P}_{3}
  2. BP1>P2>P3{P}_{1} > {P}_{2} > {P}_{3}
  3. CP2>P3>P1{P}_{2} > {P}_{3} > {P}_{1}
  4. DP3>P1>P2{P}_{3} > {P}_{1} > {P}_{2}

Correct answer

A. P2>P1>P3{P}_{2} > {P}_{1} > {P}_{3}

Step-by-step solution

Osmotic pressure (π)=CRT( \pi ) = CRT
At same temperature, π∝C\pi ∝ C

Solution: (1)( 1 )C1=10180=10180×4{C}_{1} = \dfrac{10}{180} = \dfrac{10}{180} \times 4

C1=2501000=40180=29{C}_{1} = \dfrac{250}{1000} = \dfrac{40}{180} = \dfrac{2}{9}

C1=0.222M{C}_{1} = 0 . 222M

Solution: (2)( 2 ) C2=10602501000=16×4=0.667M{C}_{2} = \dfrac{10}{\dfrac{60}{\dfrac{250}{1000}}} = \dfrac{1}{6} \times 4 = 0 . 667M

Solution: (3)( 3 ) C3=10342×1000250=0.1169M{C}_{3} = \dfrac{10}{342} \times \dfrac{1000}{250} = 0 . 1169M

∵C2>C1>C3∴∵ {C}_{2} > {C}_{1} > {C}_{3} ∴ Hence, order: P2>P1>P3{P}_{2} > {P}_{1} > {P}_{3}

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