Consider the 1M aqueous solution of the following compounds and arrange them in the increasing order of elevation in the boiling points.

A. C₆H₁₂O₆

B. NaCl

C. MgCl₂

D. AlCl₃

E. Al₂(SO₄)₃

A < B < C < D < E

The correct answer is option 3. A < B < C < D < E.

We know,

Elevation of boiling point, \(\Delta T_b = i × k_b × m\)

where,

i= van't Hoff factor

kb = molal elevation constant

m = molarity

to compare the elevation in boiling point we need to find the van't hoff factor for each compound.

A. \(C_6H_{12}O_6\) is a non-ionic compound so, its van't hoff factor \(i = 1\)

So, Elevation of boiling point, \(\Delta T_b = 1 × k_b × 1 = k_b\)

B. \(NaCl → Na^+  +  Cl^−\)

van't hoff factor \(i = 2\)

So, Elevation of boiling point, \(\Delta T_b = 2 × k_b × 1 = 2k_b\)

C. \(MgCl_2 → Mg^{2+}  +  2Cl^−\)

van't hoff factor \(i = 3\)

So, Elevation of boiling point, \(\Delta T_b = 3 × k_b × 1 = 3k_b\)

D. \(AlCl_3 → Al^{+3} + 3Cl^−\)

van't hoff factor \(i = 3\)

So, Elevation of boiling point, \(\Delta T_b = 4 × k_b × 1 = 4k_b\)

E. \(Al2(SO_4)_3 → 2Al^{+3} + 3SO_4^{−2}\)

van't hoff factor \(i = 5\)

So, Elevation of boiling point, \(\Delta T_b = 5 × k_b × 1 = 5k_b\)

So, the increasing order of elevation of boiling point ill be

\(k_b < 2k_b < 3k_b < 4k_b < 5k_b\) i.e., \(C_6H_{12}O_6 < NaCl < MgCl < AlCl_3 < Al_2(SO_4)_3\) i.e., \(A < B < C < D < E\)