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
(i) \(10^{-4}\) M NaCl
(ii) \(10^{-4}\) M Urea
(iii) \(10^{-3}\) M NaCl
(iv) \(10^{-2}\) M NaCl
Show Hint
- \((ii) <(i) <(iii) <(iv)\)
((ii) <(i) \(\sim=\) (iii) \(<(iv)\)
- \((i) <(ii) <(iii) <(iv)\)
- \((iv) <(iii) <(i) <(ii)\)
The Correct Option is A
Solution and Explanation
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:
ΔTb=iKbm
Where:
- ΔTb is the boiling point elevation,
- i is the van't Hoff factor (ion count per molecule),
- Kb 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)
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