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How would the freezing point depression ...

How would the freezing point depression of a `0.05`m `CaCl_(2)` solution compare with that of a NaCl solution ? It would be :

A

less than that for a `0.10` m NaCl solution.

B

between that for a `0.10` m NaCl solution and a `0.20` m NaCl solution .

C

between that for a `0.20` m NaCl solution and a `0.30` m NaCl solution .

D

greater than that of a `0.30` m NaCl solution .

Text Solution

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The correct Answer is:
A
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Knowledge Check

  • The freezing point of the solution M is -

    A
    268.7 K
    B
    268.5 K
    C
    234.2 K
    D
    150.9 K

  • The molecular mass of NaCl determined by studying freezing point depression of it's 0.5% aqueous solution is 30. The apparent degree of dissociantion of NaCl is :

    A
    0.95
    B
    `0.5`
    C
    `0.6`
    D
    `0.3`

  • The freezing point of water is depressed by 0.37^@ C in a 0.01 molal NaCl solution. The freezing point of 0.02 molal solution of urea is depressed by

    A
    `0.37^@C`
    B
    `0.74^@C`
    C
    `0.185^(@)C`
    D
    `0^(@)C`

    • Similar Questions

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    The freezing point depression of 0.1 molal NaCl solution is 0.372 K. What conclusion can you draw about the molecular state of NaCl in water. K_f of water = 1.86 k/m.

    Calculate the freezing point depression expected for 0.0711M aqueous solution of Na_(2) SO_(4) . If this solution actually freezes at -0.320^(@) C , what would be the value of van't Hoff factor ? (K_(f) = 1.86^(@) C mol^(-1))

    Calculate the freezing point depression expected for 0.0711 m aqueous solution of Na_(2)SO_(4) . If this solution actually freezes at -0.320 ^(@)C , what would be the value of Van't Hoff factor ? ( K_(f) for water is 1.86 ^(@)C mol^(-1) ) .

    Calculate the freezing point depression expected for 0.0711 m aqueous solution of Na_(2)SO_(4) . If this solution actually freezes at -0.320^(@)C , what would be the value of van't Hoff factor ? ( K_(f) for water is "1.86 C mol"^(-1) ).

    The freezing point depression of 0.1 m NaCl solution is 0.372^(@)C . What conclusion would you draw about its molecular state ? K_(f) for water is "1.86 K kg mol"^(-1) .

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