Question

Consider the 1M aqueous solution of the following compounds and arrange than in the increasing order of elevation in the boiling points. 
A. $C_6H_{12}O_6$ 
B. $NaCl $
C. $MgCl_2 $
D. $AlCl_3$ 
E. $Al_2(SO_4)_3 $
Choose the correct answer from the options given below: 

• A. B<C<D<E<A
• B. A<E<D<C<B
• C. A<B<C<D<E
• D. E<D<C<B<A

Answer 1

The Correct Option is A

The elevation in boiling point of a solution depends on the number of solute particles present in the solution, as well as their nature and concentration.

Let's analyze the given compounds and determine their behavior in the 1M aqueous solution:

A. $C_6H_{12}O_6$  (glucose) - Glucose is a covalent compound and does not dissociate into ions in solution. Therefore, it does not contribute additional particles to the solution.

B. $NaCl $ - NaCl dissociates into $Na^+$ and $Cl^-$ ions in solution. It forms two particles (one $Na^+$ and one $Cl^-$) for every formula unit of NaCl.

C. $MgCl_2 $ - $MgCl_2 $ dissociates into $Mg^{2+} $ and $2 Cl^-$ ions in solution. It forms three particles (one $Mg^{2+} $  and two$Cl^-$) for every formula unit of $MgCl_2 $.

D. $AlCl_3$  - $AlCl_3$  dissociates into $Al^{3+}$  and $3 Cl^-$ ions in solution. It forms four particles (one $Al^{3+}$ and three $Cl^-$) for every formula unit of $AlCl_3$.

E.  $Al_2(SO_4)_3 $-  $Al_2(SO_4)_3 $ dissociates into $2 Al^{3+}$ and $3 SO_4^{2-}$ ions in solution. It forms five particles (two $2 Al^{3+}$ and three $3 SO_4^{2-}$  for every formula unit of $Al_2(SO_4)_3 $.

Based on the number of particles formed in the solution, we can arrange the compounds in increasing order of elevation in boiling points:

B < C < D < E < A

Therefore, the correct answer is option (A) :-. B < C < D < E < A.