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

The boiling point elevation of solutions depends on solute concentration and solute type. Boiling point elevation is a colligative property, meaning it is proportional to the number of solute particles. The formula is:

ΔT_b=iK_bm

Where:

• ΔT_b represents the boiling point elevation.
• i is the van't Hoff factor, indicating the number of ions per molecule.
• K_b is the solvent's ebullioscopic constant.
• m is the solution's molality.

Analysis of each solution:

• (i) 10^-4 M NaCl: NaCl dissociates into two ions (Na⁺, Cl^-), thus i=2. For dilute solutions, molality is approximately equal to molarity.
• (ii) 10^-4 M urea: Urea does not dissociate; i=1.
• (iii) 10^-3 M NaCl: This solution has a higher concentration than (i) and i=2.
• (iv) 10^-2 M NaCl: This solution has the highest concentration and i=2.

The boiling point elevation is directly proportional to the product of i and concentration. Order the solutions by increasing boiling point elevation:

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

The resulting order of increasing boiling points is:
(ii) < (i) < (iii) < (iv)