Home
Class 12
CHEMISTRY
The benzene (V.P. = 268 mm) and ethylene...

The benzene `(V.P. = 268 mm)` and ethylene chloride `(V.P. = 236 mm)` form ideal solution. The total pressure made by dissolving 2 moles of benzene and 3 moles of ethylene chloride will be

A

504 mm

B

248.8mm

C

255.2mm

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
B

`P_("Total") = (2)/(5) xx 268 +(3)/(5) xx 236`
`= 107.2 +141.6 = 248.8mm`
Promotional Banner

Topper's Solved these Questions

  • SOLUTIONS

    DINESH PUBLICATION|Exercise REVISION QUESTION|186 Videos
  • SOLUTIONS

    DINESH PUBLICATION|Exercise OBJECTIVE TYPE MCQs|47 Videos
  • SOLID STATE

    DINESH PUBLICATION|Exercise Brain storming|10 Videos
  • SOME BASIC CONCEPTS OF CHEMISTRY

    DINESH PUBLICATION|Exercise ULTIMATE PREPARATORY PACKAGE|20 Videos

Similar Questions

Explore conceptually related problems

1 mole heptane (V.P = 92 mm of Hg) is mixed with 4 mol. Octane (V.P = 31 mm of Hg) , form an ideal solution. Find out the vapour pressure of solution.

Benzene and toluene form an ideal solution. The vapor pressures of pure benzene and pure toluene are 75 mmHg and 25 mmHg at 20^(@)C . If the mole fractions of benzene and toluene in vapor phase are 0.63 and 0.37 respectively, then the vapor pressure of the ideal mixture will be

Benzene and toluene forms an ideal solution. Vapour pressure of pure benzene is 100 torr while that of pure toluene is 50 torr. If mole faction of benzene in liquid phase is (1)/(3) . Then calculate the mole fraction of benzene in vapour phase :