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An aqueous solution of a non-electrolyte...

An aqueous solution of a non-electrolyte solute boils at `100.52^(@)C`. The freezing point of the solutiion will be (`K_(f) = 1.86 K kg "mol"^(-1), K_(b) = 0.52 K kg "mol"^(-1)`)

A

`0^(@)C`

B

`-1.86^(@)C`

C

`1.86^(@)C`

D

none of the above.

Text Solution

Verified by Experts

The correct Answer is:
B
`Delta T_(b) = K_(b) xx m`
`i=e.,0.52 = 0.52 xx m` or m=1
`Delta T_(f) =K_(f) xx m = 1.86 xx 1 = 1.86^(@)`
`i.e., T_(f) = T_(F)^(@)-Delta T_(f) = 0^(@) C-1.86^(@)C=1.86^(@)C`
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Knowledge Check

  • An aqueous solution of a non-electrolye boils at 100.52^(@)C . The freezing point of the solution will be ( K_(b) = 0.52 K kg mol^(-1), K k_(g) = 1.86 K kg mol^(-1) )

    A
    `0^(@)C `
    B
    `- 1.86^(@)C `
    C
    `1.86^(@)C `
    D
    ` - 0.52^(@)C `

  • The boiling point of an aqueous solution of a non - electrolyte is 100.52^@C . Then freezing point of this solution will be [ Given : k_f=1.86 " K kg mol"^(-1),k_b=0.52 "kg mol"^(-1) for water]

    A
    `0^@C`
    B
    `-1.86^@C`
    C
    `1.86^@C`
    D
    None of the above

  • In a 0.5 molal solution KCl, KCl is 50% dissociated. The freezing point of solution will be ( K_(f) = 1.86 K kg mol^(-1) ):

    A
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    B
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    C
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    D
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