What would be the vapour pressure of point 5 molar solution of non volatile solute in benzene at 30 degree celsius in the presence of benzene at 30 degree celsius in 119.6 star?
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The vapour pressure of pure benzene at 25∘C is 639.7mm of mercury and the vapour pressure of a solution of a solute in benzene at the same temperature is 631.9mm of mercury.Calculate the molality of the solution.
0.156mole/kg0.146mole/kg0.126mole/kg0.176mole/kg
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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