The vapor pressure of the solution of a solute in benzene is 631.9 mm of Hg at any particular temperature. Calculate the molality of the solution. (The vapor pressure of pure benzene is 639.70 mm of Hg)

Answer

0.158

Explanation

According to Raoult.s law:
`(DeltaP)/(P_A^@)=(639.7-631. 9)/(639.7) = (7.8)/(639.7) = 0.0122`
`X_B = 0.0122`
Let the total number of moles in the solution = 1
Therefore, the number of moles of solvent = 1-0.0122
` m = ("moles of solute")/("1 kg of solvent ")`
`= (0.0122)/(((1-0.0122)xx78)/(1000) )= (0.0122)/((77.05)/10000)`
`= (0.0122xx1000)/(77.05) = (12.2)/(77.05)`
`= 0.158 m`