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^(-4)` M `MgCl_(2)`
(iv)

Options

A. `(ii) lt (i) lt (iii)`
B. `(i) lt (iii) lt (ii)`
C. `(i) lt (ii) lt (iii)`
D. `(ii) lt (iii) lt (i)`

Answer

To arrange the given aqueous solutions in the order of their increasing boiling points, we need to consider the concept of colligative properties, specifically the elevation in boiling point. The boiling point elevation can be calculated using the formula:

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

Where:
- \(\Delta T_b\) = elevation in boiling point
- \(i\) = Van't Hoff factor (number of particles the solute dissociates into)
- \(K_b\) = boiling point elevation constant (which is constant for a given solvent)
- \(m\) = molality of the solution

Since all solutions have the same molality (\(10^{-4}\) M), and \(K_b\) is also constant for water, the boiling point elevation will depend solely on the Van't Hoff factor (\(i\)).

Let's analyze each solution:

1. **10^(-4) M NaCl**:
- NaCl dissociates into 2 ions: Na\(^+\) and Cl\(^-\).
- Therefore, \(i = 2\).

2. **10^(-4) M Urea**:
- Urea does not dissociate in solution; it remains as one molecule.
- Therefore, \(i = 1\).

3. **10^(-4) M MgCl\(_2\)**:
- MgCl\(_2\) dissociates into 3 ions: 1 Mg\(^{2+}\) and 2 Cl\(^-\).
- Therefore, \(i = 3\).

Now, we can summarize the Van't Hoff factors:
- NaCl: \(i = 2\)
- Urea: \(i = 1\)
- MgCl\(_2\): \(i = 3\)

Next, we can determine the order of increasing boiling points based on the Van't Hoff factors:
- The solution with the lowest \(i\) will have the lowest boiling point, and the one with the highest \(i\) will have the highest boiling point.

Thus, the order of increasing boiling points is:
1. Urea (i = 1)
2. NaCl (i = 2)
3. MgCl\(_2\) (i = 3)

**Final Answer:**
Increasing order of boiling points:
**(ii) 10^(-4) M Urea < (i) 10^(-4) M NaCl < (iii) 10^(-4) M MgCl\(_2\)**