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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

A

`0.322`

B

`0.222`

C

`0.413`

D

`0.5`

Text Solution

Verified by Experts

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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Knowledge Check

  • 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. Boiling point of the solution is

    A
    `300.73 K`
    B
    `373.165 K`
    C
    `400 K`
    D
    `273.15 K`

  • 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. Lowering in vapour pressure at 298 K is

    A
    `0.13`
    B
    `0.15`
    C
    `0.16`
    D
    `0.1378`

  • 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. Depression in freezing point of solution

    A
    `0.68`
    B
    `0.43`
    C
    `0.5989`
    D
    `0.326`

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