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The vapour pressure of pure benzene …

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The vapour pressure of pure benzene is 639.7mm of mercury and the vapour of solution of a solute in benzene at temprature is 631.9mm of mercury then find what is the molality of the solution

Posted by Jitendra Kumar 7 years, 1 month ago

CBSE > Class 12 > Chemistry

1 answers

Vineet Kumar Gupta 7 years, 1 month ago

Let the molality of the solution =m Now the solution contains m moles of solute per 1000gm of benzene. Vapour pressure of pure benzene,P0=639.7mm Vapour pressure of solution ,P=631.9mm Moles of benzene (mol wt. 78)N=100078 Moles of solute,n=? Substituting these values in the Raoult's equations P0−PP0=nN 639.7−631.9639.7=n×781000 (Or) 7.8639.7=78n1000 ∴n=1000×7.878×639.7 ⇒0.156 Hence molality of solution =0.156mole/kg

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