Question

The solution is prepared by mixing 8.5 g of CH2Cl2 and 11.95 g of CHCl3 at 298 K are 415 and 200 mmHg respectively, the mole fraction of CHCl3 in vapour form is: (Molar mass of Cl=35.5 g molâˆ’1) (JEE Main - 2017)

A
0.325
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B
0.486
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C
0.162
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D
0.675
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Solution

The correct option is A 0.325Molar mass of CHCl3=119.5 g/mole Molar mass of CH2Cl2=85 g/mole Moles of CHCl3=11.95119.5=0.1 mole Moles of CH2Cl2=8.585=0.1 moles Mole fraction of CHCl3=0.10.2=0.5 mole Mole fraction of CH2Cl2=0.10.2=0.5 mole Given: Vapour pressure of CHCl3=200 mmHg=0.263 atm Vapour pressure of CH2Cl2=415 mmHg=0.546 atm [∵1 atm=760 mmHg] Pabove solution=Mole fraction of CHCl3×vapour pressure of CHCl3+Mole fraction of CH2Cl2×vapour pressure of CH2Cl2 =0.5×0.263+0.5×0.546 =0.4045 Mole fraction of CHCl3 in vapour form=0.13150.4045=0.325

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