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JEE Main 202117 MarchMorning Shift
Exam
JEE Main
Year
2021
Date
17 March
Shift
Morning Shift
Subject
Chemistry
Chapter
Some Basic Concepts of Chemistry

JEE Main 2021 · 17 March · Morning Shift · Chemistry · Some Basic Concepts of Chemistry

The mole fraction of a solute in a 100 molal aqueous solution...

The mole fraction of a solute in a 100100 molal aqueous solution ________ ×10−2.\times {10}^{- 2} .

(Round off to the Nearest Integer).

[Given : Atomic masses: H:1.0u,O:16.0uH: 1 . 0u, O: 16 . 0u]

Correct answer

0

Step-by-step solution

100100 molal aqueous solution means there is 100100 mole solute in 1kg=1000gm1kg = 1000gm water.

Now,

mole-fraction of solute=nsolutensolute+nsolvent= \dfrac{{n}_{\text{solute}}}{{n}_{\text{solute}} + {n}_{\text{solvent}}}

=100100+100018=18002800=0.6428= \dfrac{100}{100 + \dfrac{1000}{18}} = \dfrac{1800}{2800} = 0 . 6428

=64.28×10−2= 64 . 28 \times {10}^{- 2}

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