Question

Arrange the following solutions in order of their increasing boiling points.
(i) \(10^{-4}\) M NaCl
(ii) \(10^{-4}\) M Urea
(iii) \(10^{-3}\) M NaCl
(iv) \(10^{-2}\) M NaCl

• A. \((ii) <(i) <(iii) <(iv)\)
• B. \((ii) <(i) \(\sim=\)\) (iii) \(<(iv)\)
• C. \((i) <(ii) <(iii) <(iv)\)
• D. \((iv) <(iii) <(i) <(ii)\)

Hint:

For colligative properties like boiling point elevation, the more the solute particles (ions), the higher the boiling point. Solutions with higher concentrations or higher ionization will have higher boiling points.

Answer 1

The Correct Option is A

To solve this problem, we need to understand how the boiling point of solutions is affected by their concentration and the nature of the solute. The boiling point elevation is a colligative property, which means it depends on the number of solute particles in a solution. The formula for boiling point elevation is:

ΔT_b=iK_bm

Where:

• ΔT_b is the boiling point elevation,
• i is the van't Hoff factor (ion count per molecule),
• K_b is the ebullioscopic constant of the solvent,
• m is the molality of the solution.

Let's analyze each solution:

• (i) 10^-4 M NaCl: NaCl dissociates into 2 ions (Na⁺, Cl^-), so i=2. Molality directly correlates to molarity due to its dilute solution.
• (ii) 10^-4 M urea: Urea does not dissociate; i=1.
• (iii) 10^-3 M NaCl: Higher concentration than (i) with i=2.
• (iv) 10^-2 M NaCl: Highest concentration with i=2.

Order these solutions by their boiling point elevation. Higher the product of i and concentration, higher the boiling point:

Solution	i	Concentration	i×Concentration
(i) 10-4 M NaCl	2	10-4	2×10-4
(ii) 10-4 M Urea	1	10-4	1×10-4
(iii) 10-3 M NaCl	2	10-3	2×10-3
(iv) 10-2 M NaCl	2	10-2	2×10-2

Resulting order of increasing boiling points:
(ii) < (i) < (iii) < (iv)