A) 0.1 M glucose
B) \[0:1\,M\,BaC{{l}_{2}}\]
C) \[0.1\,M\,MgS{{O}_{4}}\]
D) \[0.1\text{ }NaCl\]
Correct Answer: B
Solution :
Lowering of vapour pressure is a colligative property ie, depends only upon the number of particles of solute and not on the nature of solute. \[\because \]0.1 M Glucose\[\xrightarrow{{}}\]remains undissociated \[0.1\,M\text{ }BaC{{l}_{2}}\xrightarrow{{}}B{{a}^{2+}}+2C{{l}^{-}}\Rightarrow 3\text{ }ions\] \[0.1\,M\,MgS{{O}_{4}}\xrightarrow{{}}M{{g}^{2+}}+SO_{4}^{2-}\Rightarrow 2\,ions\] \[0.1\,M\,NaCl\xrightarrow[{}]{{}}N{{a}^{+}}+C{{l}^{-}}\Rightarrow 2\,ions\] \[\therefore \]\[0.1\text{ }M\text{ }BaC{{l}_{2}}\] gives maximum number of particles, hence it exhibits maximum lowering of vapour pressure.
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