JEE Main 9 April 2017 Paper Solved Paper

Section: Chemistry

Question : 40 of 90

Marks: +1, -0

A solution is prepared by mixing 8.5 g of CH2Cl2 and 11.95 g of CHCl3. If vapour pressure of CH2Cl2 and CHCl3 at 298 Kare 415 and 200 mmHg respectively, the mole fraction of CHCl3 in vapour form is :(Molar mass of Cl=35.5gmol−1

[Main 9 April 2017]

0.162
0.675
0.325
0.486

Validate

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