Which of the following is correct increasing order of freezing point for these four solutions?

$I . 0.1 M$ glucose $I I .0 .1 M$ ${N a}_{3} {P O}_{4}$
$I I I .0 .1 M$ $N a C l$ $I V .0 .1 M$ $K_{4} [F e {(C N)}_{6}]$

I < I I < I I I < I V

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I < I I I < I I < I V

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Solution

The correct option is C $I V < I I < I I I < I$
Greater the number of species, higher the depression in freezing point and lower the freezing point.
Hence, Freezing point $\propto \frac{1}{i}$

Glucose, $= 1;$ Since it is a non-electrolyte.

$N a_{3} P O_{4} ⟶ 3 N a^{+} + P O_{4}^{3 -}$
$i = 4;$

$N a C l ⟶ N a^{+} + C l^{-}$
$i = 2;$

$K_{4} {[F e {(C N)}_{6}] ⟶ 4 K^{+} + [F e (C N)_{6}]}_{6}^{4 -}$
$i = 5$

Hence, the correct answer is option B.

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The correct order of this equivalent conductances

Which of the following options represent their correct order of equivalent conductance?

[F e (C N)_{6}]^{4 -}

[F e (C N)_{6}]^{3 -}

[C r (N H_{3})_{6}]^{3 +}

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[F e (C N)_{6}]^{4 -}

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[C r (N H_{3})_{6}]^{3 +}

[N i (H_{2} O)_{6}]^{2 +}

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