Which of the following options are correctly representing the increasing order of freezing points of aqueous solutions?

0.01 m A l_{2} (S O_{4})_{3} > 0.01 m B a C l_{2} > 0.01 m N a C l > 0. 01 m g l u c o s e

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0.01 m B a C l_{2} > 0.01 m N a C l > 0.01 m C H_{3} C O O H

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0.01 m g l u c o s e > 0.01 m C H_{3} C O O H > 0.01 m K C l

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0.01 m B a C l_{2} = 0.01 m K C l = 0.01 m G l u c o s e

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Solution

The correct option is C $0.01 m g l u c o s e > 0.01 m C H_{3} C O O H > 0.01 m K C l$
We know, $Δ T_{b} = i K_{f} m$
Here, $Δ T_{b}$ = depression of freezing point
$Δ T_{b} = freezing point of the pure solvent - freezing poin of the solution$

$∴ Freezing point of solution \propto \frac{1}{D . F . P}$
D.F.P. = Depression in freezing
Again, $D . F . P . \propto i \times m$
Where,
i = van't Hoff factor
and m = molality
Now,
$A l_{2} (S O_{4})_{3} \to 2 A l^{3 +} + 3 S O_{4}^{2 -} i_{A l_{2} (S O_{4})_{3}} = 5 B a C l_{2} \to B a^{2 +} + 2 C l^{-} i_{B a C l_{2}} = 3 N a C l \to N a^{+} + C l^{-} i_{N a C l} = 2 i_{glucose} = 1 K C l \to K^{+} + C l^{-} i_{K C l} = 2 i_{C H_{3} - C O O H} = greater than 1 but less than 2$
because $C H_{3} - C O O H$ does not undergo 100% ionization.

Here, molality for all given solution are same. So depression in freezing (DFP) depend only on the value of 'i'
$∴ Increasing order of D.F.P can be given as$
$0.01 m g l u c o s e < 0.01 m C H_{3} - C O O H < 0.01 m K C l < 0.01 m B a C l_{2} < 0.01 m A l_{2} (S O_{4})_{3}$
Hence,
$Decreasing order of freezing point$
$0.01 m glucose > 0.01 m C H_{3} - C O O H > 0.01 m K C l$

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