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Vapour pressure of CH(3)Cl and CH(2)Cl(2...

Vapour pressure of `CH_(3)Cl` and `CH_(2)Cl_(2)` are 540 mm Hg and 402 mm Hg respectively. 101 g of `CH_(3)Cl` and 85 g of `CH_(2)Cl_(2)` are mixed together. Determine
(i) The pressure at which the solution starts boiling.
(ii) Molar ratio of solute v/s solvent in vapour phase in equilibrium with solution.

Text Solution

Verified by Experts

Boiling occurs when external pressure becomes equal to the vapour pressure. So, the boiling pressure= V.P. of solution
`=p_A^@ x_A+p_B^@+x_B`
Let A=`CH_3Cl,B=CH_2 Cl_2`, then
Total pressure =`540 times 2/3+402 times 1/3`
=360+134=494 mm Hg
here the solute is `CH_2Cl_2` as mass is less
`x._(Ch_2Cl_2)=(402 times 1/3)/494=134/494`
`x._(CH_3Cl)=(540 times 2/3)/494=360/494`
Now, `(n._(CH_2Cl_2))/(n.(CHCl_3))=n_(solution g)/n_(solvent g)=134/360=0.372`
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