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Freezing point of urea solution is -0.6^...

Freezing point of urea solution is `-0.6^(0)C`. How much urea is required to be dissolved in 3 kg water ? `[M("urea")=60 g mol^(-1), K_(f)=1.5^(0)"C Kg mol"^(-1)]`

A

2.4 g

B

3.6 g

C

6.0 g

D

72 g

Text Solution

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The correct Answer is:
D
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Freezing point of urea solution is -0.6^@C .(Kf = 1.50 C Kg/mol) How much urea (M.W.= 60 gm/mole) required to dissolved in 3 kg water ?

The aqueous solution of urea has freezing point -0.6^(@)C . To prepare such solution how much gram of urea is needed to dissolve in 3 kg of water ? (M = 60 gm/mol) (K_(f)=1.5^(@)C" kg mol"^(-1)) .

Knowledge Check

  • The freezing point of the aqueous solution of urea is -0.6^(@)C . How much urea should be added to 3 kg water to get such solution ? (K_(f)=1.5^(@)"C Kg. mol"^(-1))

    A
    72 gram
    B
    6.0 gram
    C
    3.6 gram
    D
    2.4 gram

  • To produce difference between freezing point and boiling point of a solution by 105.0^(@)C , how much sucrose should be dissolved in 100 gm of water ? (K_(f)=1.86^(@)C, kg mol^(-1) " and " K_(b)=0.151^(@)Ckg mol^(-1))

    A
    72 gm
    B
    34.2 gm
    C
    342 gm
    D
    460 gm

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