1.

Calculate the freezing point of an aqueous soltuion of non-electrolyte having an osmotic pressure 2.0 atm at 300 K. (K'_(f) = 1.86 K mol^(-1) kg and S = 0.0821 litre atm K^(-1) mol^(-1))

Answer»

`-0.15^(@)C`
`+0.15^(@)C`
`-0.51^(@)C`
`-0.17^(@)C`

SOLUTION :`PI = CRT`
`:. C = (pi)/(RT)`
`=(2.0 atm)/((0.0821 L atm K^(-1) MOL^(-1))xx(300K))`
`=0.0812 mol L^(-1) =0.0812 M ~~0.0812 m`
`( :'` solution is very dilute)
`DELTA T_(f) = K_(f)m = (1.86 Km^(-1)) xx (0.0812m)`
`=0.15K = 0.15^(@)C`
`:.` Freezing point `= (0.00 - 0.15)^(@)C =- 0.15^(@)C`.


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