Arrange the following aqueous solutions in the order of their increasing boiling points
(i)`10^(-4)` M NaCl , (ii)`10^(-4)` M Urea , (iii)`10^(-3)` M `MgCl_2` , (iv)`10^(-2)` M NaCl

Options

A. (i) lt (ii) lt (iv) lt (iii)
B. (ii) lt (i) = (iii) lt (iv)
C. (ii) lt (i) lt (iii) lt (iv)
D. (iv) lt (iii) lt (i) = (ii)

Explanation

To arrange the given aqueous solutions in the order of their increasing boiling points, we will use the concept of boiling point elevation, which is influenced by the van 't Hoff factor (i) and the molarity (M) of the solutions. The formula for boiling point elevation is given by:

\[ \Delta T_b = i \cdot K_b \cdot m \]

Where:
- \(\Delta T_b\) = boiling point elevation
- \(i\) = van 't Hoff factor (number of particles the solute dissociates into)
- \(K_b\) = ebullioscopic constant (which is the same for all solutions)
- \(m\) = molality (which is proportional to molarity for dilute solutions)

Since \(K_b\) is constant for all solutions, we can focus on the product \(i \cdot M\) to determine the boiling point elevation for each solution.

### Step-by-step solution:

1. **Calculate \(i \cdot M\) for each solution:**
- **(i)** \(10^{-4} \, M \, \text{NaCl}\):
- \(i = 2\) (NaCl dissociates into Na\(^+\) and Cl\(^-\))
- \(i \cdot M = 2 \cdot 10^{-4} = 2 \times 10^{-4}\)

- **(ii)** \(10^{-4} \, M \, \text{Urea}\):
- \(i = 1\) (Urea does not dissociate)
- \(i \cdot M = 1 \cdot 10^{-4} = 1 \times 10^{-4}\)

- **(iii)** \(10^{-3} \, M \, \text{MgCl}_2\):
- \(i = 3\) (MgCl\(_2\) dissociates into Mg\(^{2+}\) and 2 Cl\(^-\))
- \(i \cdot M = 3 \cdot 10^{-3} = 3 \times 10^{-3}\)

- **(iv)** \(10^{-2} \, M \, \text{NaCl}\):
- \(i = 2\) (NaCl dissociates into Na\(^+\) and Cl\(^-\))
- \(i \cdot M = 2 \cdot 10^{-2} = 2 \times 10^{-2}\)

2. **List the calculated values:**
- (i) \(2 \times 10^{-4}\)
- (ii) \(1 \times 10^{-4}\)
- (iii) \(3 \times 10^{-3}\)
- (iv) \(2 \times 10^{-2}\)

3. **Order the solutions based on \(i \cdot M\) values:**
- The smallest value is \(1 \times 10^{-4}\) (Urea).
- Next is \(2 \times 10^{-4}\) (NaCl).
- Then \(3 \times 10^{-3}\) (MgCl\(_2\)).
- Finally, \(2 \times 10^{-2}\) (NaCl).

4. **Arrange in order of increasing boiling points:**
- (ii) \(10^{-4} \, M \, \text{Urea}\)
- (i) \(10^{-4} \, M \, \text{NaCl}\)
- (iii) \(10^{-3} \, M \, \text{MgCl}_2\)
- (iv) \(10^{-2} \, M \, \text{NaCl}\)

### Final Answer:
The order of increasing boiling points is:
\[ \text{Urea} < \text{NaCl} < \text{MgCl}_2 < \text{NaCl} \]