MCQ
For $0.1\,M $ solution, the colligative property will follow the order
- A$NaCl > N{a_2}S{O_4} > N{a_3}P{O_4}$
- ✓$NaCl < N{a_2}S{O_4} < N{a_3}P{O_4}$
- C$NaCl > N{a_2}S{O_4}\,\, \approx \,\,N{a_3}P{O_4}$
- D$NaCl < N{a_2}S{O_4} = N{a_3}P{O_4}$
✓
Answer
Correct option: B.
$NaCl < N{a_2}S{O_4} < N{a_3}P{O_4}$
b
(b)Colligative property in decreasing order
(b)Colligative property in decreasing order
$N{a_3}P{O_4} > N{a_2}S{O_4} > NaCl$
$N{a_3}P{O_4} \to 3N{a^ + } + PO_4^{3 - } = 4$
$N{a_2}S{O_4} \to 2N{a^ + } + SO_4^{2 - } = 3$
$NaCl \to N{a^ + } + C{l^ - } = 2$
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