For $0.1 \mathrm{~M}$ solution, the colligative property will follow the order

For $0.1 \mathrm{~M}$ solution, the colligative property will follow the order
  1. $\mathrm{NaCl}>\mathrm{Na}_{2} \mathrm{SO}_{4}>\mathrm{Na}_{3} \mathrm{PO}_{4}$
  2. $\mathrm{NaCl} < \mathrm{Na}_{2} \mathrm{SO}_{4} < \mathrm{Na}_{3} \mathrm{PO}_{4}$
  3. $\mathrm{NaCl}>\mathrm{Na}_{2} \mathrm{SO}_{4} \approx \mathrm{Na}_{3} \mathrm{PO}_{4}$
  4. $\mathrm{NaCl} < \mathrm{Na}_{2} \mathrm{SO}_{4}=\mathrm{Na}_{3} \mathrm{PO}_{4}$

Solution

Colligative property in decreasing order
$\mathrm{Na}_{3} \mathrm{PO}_{4}>\mathrm{Na}_{2} \mathrm{SO}_{4}>\mathrm{NaCl}$
$\mathrm{Na}_{3} \mathrm{PO}_{4} ightarrow 3 \mathrm{Na}^{+}+\mathrm{PO}_{4}^{3-}=4$
$\mathrm{Na}_{2} \mathrm{SO}_{4} ightarrow 2 \mathrm{Na}^{+}+S O_{4}^{2-}=3$
$\mathrm{NaCl} ightarrow \mathrm{Na}^{+}+\mathrm{Cl}^{-}=2$ *

Asked in: JEE-TOPICTESTS-CHEMISTRY

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