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The freezing point of a solution of 2.40...

The freezing point of a solution of 2.40 g of biphenyl(`C_(12)H_(10)`) in 75.0 g of benzene (`C_(6)H_(6)`) is `4.40^(@)C`. The normal freezing point of benzene is `5.50^(@)C`. What is the molal freezing point constant `(^(@)C//m)` for benzene ?

A

5.3

B

5.1

C

4.6

D

4.8

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The correct Answer is:
a
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Knowledge Check

  • The molal freezing point constant for water is 1.86^(@)C//m . Therefore, the freezing point of 0.1 M

    A
    `-272.628K`
    B
    `+272.628K`
    C
    `-0.372^(@)C`
    D
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  • What is the freezing point of a solution contains 10.0g of glucose C_(6)H_(12)O_(6) , in 100g of H_(2)O ? K_(f)=1.86^(@)C//m

    A
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    B
    `+0.186^(@)C`
    C
    `-0.10^(@)C`
    D
    `-1.03^(@)C`

  • The freezing point of a solution containing 4.8 g of a compound in 60 g of benzene is 4.48. What is the molar mass of the compound (K_(f)=5.1 km^(-1)) , (freezing point of benzene = 5.5^(@)C )

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    B
    200
    C
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    D
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