Calculate the freezing point of an aqueo...

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)`)

Text Solution

Verified by Experts

We know that,
P = CRT
or `C=(P)/(RT)=(2)/(0.0821xx300)"mol lit"^(-1)`
In dilute solution, the density of water can be taken as `1.0g cm^(-3)`.
Hence, `(2)/(0.0821xx300)"mol lit"^(-1)=(2)/(0.081xx300)"mol kg"^(-1)`
Let `Delta T` be the depression in freezing point
`DeltaT = K_(f)xx` molality
`= 1.86xx(2)/(0.821xx300)=0.151K`
Freezing point `=(273 -0.151) = 272.49 K or - 0.151^(@)C`.

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The freezing point of an aqueous solution of non-electrolyte having an osmotic pressure of 2.0 atm at 300 K is (Given K_(f) = 1.86 km^(-1) and R = 0.0821 L atm K^(-1) mol^(-1))

`-0.15^(@)C`

`+0.15^(@)C`

`-0.51^(@)C`

`-0.17^(@)C`

The freezing point of an aqueous solution of non-electrolyte having an osmotic pressure of 2.0 atm at 300 K will be, (K_(f) "for " H_(2)O = 1.86K mol^(-1)g) and R = 0.0821 litre atm K^(-1) mol^(-1) . Assume molarity and molarity to be same :

`-0.151^(@)C`

`-1.151^(@)C`

`-3.151^(@)C`

`-2.151^(@)C`

The freezing point of an aqueous solution of non-electrolyte having an osmotic pressure of 2.0 atm at 300 K will be, (K_(f) "for " H_(2)O = 1.86K mol^(-1)g) and R = 0.0821 litre atm K^(-1) mol^(-1) . Assume molarity and molality to be same :