Question

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

$I.0.1M$ glucose $II.0.1M$ ${Na}_3{PO}_4$
$III.0.1M$ $NaCl$ $IV.0.1M$ $K_4\left[Fe{\left(CN\right)}_6\right]$

• A. $I<II<III<IV$
• B. $IV<II<III<I$
• C. $I<III<II<IV$
• D. $IV<I<III<II$

Answer 1

Answer: (B)

$IV<II<III<I$

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.

$N{a}_{3}P{O}_4⟶3N{a}^{+}+P{O}_4^{3-}$

$i=4;$

$NaCl⟶N{a}^{+}+C{l}^{-}$

$i=2;$

$K_4{\left[Fe{\left(CN\right)}_6\right]⟶4K^{+}+\left[Fe\right(CN{)}_6]}^{4-}$

$i=5$

Hence, the correct answer is option B.