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JEE Main 20206 SeptemberEvening Shift
Exam
JEE Main
Year
2020
Date
6 September
Shift
Evening Shift
Subject
Chemistry
Chapter
Solutions

JEE Main 2020 · 6 September · Evening Shift · Chemistry · Solutions

A set of solutions is prepared using 180g of water as a solvent and...

A set of solutions is prepared using 180g180g of water as a solvent and 10g10g of different non-volatile solutes A,BA, B and CC. The relative lowering of vapour pressure in the presence of these solutes are in the order [Given, molar mass of A=100gmol−1;B=200gmol−1;C=10,000gmol−1A = 100g{mol}^{- 1}; B = 200g{mol}^{- 1}; C = 10, 000g{mol}^{- 1}]

Options

  1. AB>C>AB > C > A
  2. BC>B>AC > B > A
  3. CA>B>CA > B > C
  4. DA>C>BA > C > B

Correct answer

C. A>B>CA > B > C

Step-by-step solution

Number of moles = wt(gm)MW=10MW\dfrac{wt\left(gm\right)}{MW} = \dfrac{10}{MW}

Relative lowering in vapour pressure depends on no. of mole of solute greater the no. of moleof solute greater in RLVP and smaller will be vapour pressure.

So order of vapour pressure is A>B>CA > B > C.

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