Consider the following aqueous solutions.I. 2.2 g Glucose in 125 mL of solution.II. 1.9 g Calcium ch

Question

Consider the following aqueous solutions.

I. 2.2 g Glucose in 125 mL of solution.

II. 1.9 g Calcium chloride in 250 mL of solution.

III. 9.0 g Urea in 500 mL of solution.

IV. 20.5 g Aluminium sulphate in 750 mL of solution.

The correct increasing order of boiling point of these solutions will be:

[Given: Molar mass in $g{mol}^{- 1}:H = 1,C = 12,\text{ }N = 14$, $O = 16,Cl = 35.5,Ca = 40,Al = 27$ and $S = 32$ ]

Accepted Answer

$\ \Delta T_{b} = i \cdot k_{b} \cdot m$

For dilute solution ( $M = m$ )

Molarity	$$\mathbf{i \times m}$$
(I) $M_{\text{glucose~}} = \frac{2.2}{180} \times \frac{1000}{125} = 0.098$	$$0.098 \times 1$$
(II) $M_{{CaCl}_{2}} = \frac{1.9}{111} \times \frac{1000}{250} = 0.068$	$$0.068 \times 3$$
(III) $M_{\text{urea~}} = \frac{9}{60} \times \frac{1000}{500} = 0.3$	$$0.3 \times 1$$
(IV) $M_{{Al}_{2}\left( {SO}_{4} \right)_{2}} = \frac{20.5}{342} \times \frac{1000}{750} \simeq 0.08$	$$0.08 \times 5$$

Order of $\Delta T_{b} = {Al}_{2}\left( {SO}_{4} \right)_{3} >$ Urea $> {CaCl}_{2} >$ Glucose

So order of $BP = {Al}_{2}\left( {SO}_{4} \right)_{3} >$ Urea $> {CaCl}_{2} >$ Glucose

So Answer will be I $<$ II $<$ III $<$ IV

Question