AMU Medical AMU Solved Paper-2000

The freezing points of equimolar solutions of glucose, \[KN{{O}_{3}}\] and \[AlC{{l}_{3}}\] are in the order of

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

Greater the value of i, greater the value of \[\Delta {{T}_{f}}\] but smaller the value of freezing point. Therefore, the freezing points of equimolar solutions of glucose, \[KN{{O}_{3}}\] and \[AlC{{l}_{3}}\] are in the order of \[AlC{{l}_{3}}<KN{{O}_{3}}<\] glucose

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AMU Medical AMU Solved Paper-2000

question_answer The freezing points of equimolar solutions of glucose, \[KN{{O}_{3}}\] and \[AlC{{l}_{3}}\] are in the order of 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 \[<AlC{{l}_{3}}<KN{{O}_{3}}\]

The freezing points of equimolar solutions of glucose, \[KN{{O}_{3}}\] and \[AlC{{l}_{3}}\] are in the order of

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 \[<AlC{{l}_{3}}<KN{{O}_{3}}\]