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An aqueous solution freezes at 272.4 K w...

An aqueous solution freezes at 272.4 K while pure water freezes at 273 K. Given `K_(f)=1.86 K kg "mol"^(-1)`,`K_(b)=0.512 K kg "mol"^(-1)` and vapour pressure of water at 298 K = 23.756 mm Hg. Determine the following.
Molality of the solution is

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The correct Answer is:
A
Molarity of solution
`DeltaT_(f)=K_(f) xx m rArr =(DeltaT_(f))/(K_(f))`
`DeltaT_(f)=273-272.4 = 0.6K`,`K_(f)=0.86 K kg mol^(-1)`
`:.m=0.6/1.86=0.322`
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