Question

Which of the following has highest boiling point.
A)0.1 M nacl
B)0.1 M glucose C)0.1M Al2(so4)3
D) 0.1M cacl2

Answer 1

Answer:

C) Al2(SO4)3

Explanation:

As We Know That Here Dissociation take place leads to Abnormal Molar Mass Which Means Van't Hoff Factor (i)  Comes In So the Formula Of Elevation In Boiling Point Modified into:-

Δ$Tb = i m Kb$.

For Al2(SO4)3 i =5 (highest among all)

Answer 2

Answer:

0.1M Al₂(SO₄)₃ has highest boiling point .

Therefore option (C) is correct.

Explanation:

We know,  the elevation in boiling point is :

        Δ$T_{b}=i.m.k_{b}$

where $i$ is van't hoff factor

          $m$ is molality of solution

        $k_{b}$ is molal boiling point constant.

Elevation in boiling point is directly proportional to number of ions in the solution i.e. vant' hoff factor since the concentration is constant.

• NaCl dissociates  in aqueous solution as:

                     $NaCl$  →  $Na^{+}+Cl^{-}$

        Number of ions of $NaCl$ is two ions.

• Glucose will not dissociates in the aqueous solution.
• $Al_{2}(SO_{4})_{3}$ dissociates into 5 ions in aqueous solution.

             $Al_{2}(SO_{4})_{3}$    →   $2Al^{3+}$  +  $3SO_{4}^{2-}$

• CaCl₂ dissociates into three ions in aqueous solution.

                         $CaCl_{2}$  →   $Ca^{2+}+2Cl^{-}$

Therefore Al₂(SO₄)₃ has highest boiling point because it has highest number of solute particles in aqueous solution.