A) \[AlC{{l}_{3}}<\] glucose \[<KN{{O}_{3}}\]
B) \[AlC{{l}_{3}}<KN{{O}_{3}}<\] glucose
C) glucose \[<KN{{O}_{3}}<AlC{{l}_{3}}\]
D) glucose \[<AlC{{l}_{3}}<KN{{O}_{3}}\]
Correct Answer: B
Solution :
Key Idea \[T={{T}_{0}}-(\Delta {{T}_{f}})\] where, \[{{T}_{0}}\] is freezing temperature of solvent \[\Delta {{T}_{f}}=i\times {{K}_{f}}\times m\] where, m is molality, \[{{K}_{f}}\] is molal depression constant i = vant Hoff factor \[i=[1+(y-1)\,x]\] where y = number of products (ion or molecule) obtained per mole of the reactant. Here the solutions are equimolar. So, \[\Delta {{T}_{f}}\]depend upon i.Solute | Ionisation product | Y | \[i=[1+(y-1)x]\]her , \[x=1\] |
Glucose | No (it is non-electrolvte) | 1 | 1 |
\[KN{{O}_{3}}\] | \[{{K}^{+}}+NO_{3}^{-}\] | 2 | 2 |
\[AlC{{l}_{3}}\] | \[A{{l}^{3+}}+3C{{l}^{-}}\] | 4 | 4 |
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