A) Urea
B) \[0.1M\] glucose
C) \[0.1\,M\,MgS{{O}_{4}}\]
D) \[0.1\,M\,BaC{{l}_{2}}\]
Correct Answer: D
Solution :
\[\frac{{{P}^{0}}-{{P}_{s}}}{{{P}^{0}}}=\text{molality}\,\times \text{(1-}\alpha +x\alpha +y\alpha )\] the value of \[{{P}^{0}}-{{P}_{s}}\] is maximum for \[BaC{{l}_{2}}\].
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