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 (1) (ii) < (i) < (iii) < (iv) (2) (ii) < (i)  (iii) < (iv) (3) (i) < (ii) < (iii) < (iv) (4) (iv) < (iii) < (i) < (ii)

🔍 Key Concepts

1. Boiling-point elevation (a colligative property)
   The rise in boiling point $\Delta T_b$ depends only on number of solute particles in solution, not on their identity.

• $K_b$ – molal elevation constant (fixed for the solvent, here water).
• $m$ – molality (almost same as molarity for very dilute solutions).
• $i$ – van’t Hoff factor (actual particle count per formula unit).

2. Electrolyte vs. non-electrolyte

   • Urea does not ionise: $i = 1$.
   • NaCl dissociates: $\text{NaCl} \to \text{Na}^+ + \text{Cl}^- \ \Rightarrow\ i \approx 2$.

3. Relative comparison
   Since $K_b$ is common, compare $i\,m$ only.

Smaller $i m$ → smaller $\Delta T_b$ → lower actual boiling point.

Logical Steps a Student Follows

1. Identify type of solute (electrolyte vs. non-electrolyte).
2. Assign van’t Hoff factor $i$.
3. Multiply $i$ by given molarity (treating as molality for dilute water solutions).
4. Rank solutions in ascending order of $i m$.
5. Map that ranking directly to ascending boiling points.

Thus,

Leading to the sequence (ii) < (i) < (iii) < (iv) (option 1).

🧒 Imagine Soup Getting Hotter

1. Boiling point is the temperature where a liquid starts turning into steam.
2. When you dissolve something in water, it is like adding tiny obstacles for water molecules to escape. More obstacles → water needs more heat → higher boiling point.
3. Two things decide how many obstacles there are:
   • How many particles actually float around (1 lump of sugar → 1 particle; 1 grain of salt → splits into 2 particles, Na⁺ and Cl⁻).
   • How many lumps you added (concentration).
4. So, we look at number of particles × concentration.
5. In the list:
   • Urea stays whole (1 particle) and is tiny in amount.
   • NaCl splits into 2 particles and appears in different amounts.
6. Fewer particles → lower boiling point rise. So the order from lowest to highest is:

Urea (ii) → Dilute NaCl (i) → Medium NaCl (iii) → Strong NaCl (iv)

Hence option (1) is correct.

Step-by-Step Calculation

1. Write the formula

Since all solutions are extremely dilute and in the same solvent, compare only $i m$.

2. Calculate $i m$ for each solution

• (ii) 10(^{-4}) M Urea:
  $i = 1$ (non-electrolyte)
  $i m = 1 \times 10^{-4} = 1.0 \times 10^{-4}$

• (i) 10(^{-4}) M NaCl:
  $i \approx 2$
  $i m = 2 \times 10^{-4} = 2.0 \times 10^{-4}$

• (iii) 10(^{-3}) M NaCl:
  $i \approx 2$
  $i m = 2 \times 10^{-3} = 2.0 \times 10^{-3}$

• (iv) 10(^{-2}) M NaCl:
  $i \approx 2$
  $i m = 2 \times 10^{-2} = 2.0 \times 10^{-2}$

3. Rank $i m$ values (smallest → largest)

4. Translate to boiling points
   Lower $i m$ → smaller $\Delta T_b$ → lower boiling point.

Hence, increasing boiling point order:

5. Select the correct option
   This matches Option (1).

Final answer: Option (1).