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An aqueous solution at -2.55^(@)C. What ...

An aqueous solution at `-2.55^(@)C`. What is its boiling point `(K_(b)^(H_(2)O)=0.52 K m^(-1)`,`K_(f)^(H_(2)O)=1.86 K m^(-1)`?

A

`107.0^(@)C`

B

`100.6^(@)C`

C

`100.1^(@)C`

D

`100.7^(@)C`

Text Solution

Verified by Experts

The correct Answer is:
D
`DeltaT_(f)=2.55^(@)C = K_(f).m :. M=2.55/K_(f)`
`DeltaT_(b)=K_(b)m rArr DeltaT_(b)`
`=2.55 xx K_(b)/K_(f)=2.55 xx 0.52/1.86^(@)C)=0.7^(@)C`
`rArr T_(b) = 100 + 0.7 =100.7^(@)C`
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Knowledge Check

  • An aqueous solution of NaCI freezes at -0.186^(@)C . Given that K_(b(H_(2)O)) = 0.512K kg mol^(-1) and K_(f(H_(2)O)) = 1.86K kg mol^(-1) , the elevation in boiling point of this solution is:

    A
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    B
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    C
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    D
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    A
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    B
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    D
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  • Elevation in boiling point of an aqueous solution of a non-electroluyte solute is 2.01^(@) . What is the depression in freezing point of the solution ? K_(b)(H_(2)O)=0.52^(@)mol^(-1)kg K_(f)(H_(2)O)=1.86^(@)mol^(-1)kg

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    B
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    C
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