The osmotic pressure of equimolar solutions of glucose, `KCI`, `MgCl_2` follows the order

Options

A. `MgCl_2 gt KCI gt Glucose`
B. `Glucose gt KCI gt MgCl_2`
C. `KCI gt MgCl_2 gt Glucose`
D. `MgCl_2 gt Glucose gtKCl`

Answer

To determine the order of osmotic pressure of equimolar solutions of glucose, KCl, and MgCl₂, we need to consider the van 't Hoff factor (i) for each solute. The osmotic pressure (π) is given by the formula:

\[ \pi = iCRT \]

Where:
- \( \pi \) = osmotic pressure
- \( i \) = van 't Hoff factor (number of particles the solute dissociates into)
- \( C \) = molarity of the solution
- \( R \) = ideal gas constant
- \( T \) = temperature in Kelvin

Since we are dealing with equimolar solutions, the concentration (C) and temperature (T) are constant for all three solutions. Therefore, the osmotic pressure will depend primarily on the van 't Hoff factor (i).

### Step-by-Step Solution:

1. **Identify the van 't Hoff factor (i) for each solute:**
- **Glucose (C₆H₁₂O₆)**: Glucose is a non-electrolyte and does not dissociate in solution. Therefore, \( i = 1 \).
- **KCl**: KCl is a strong electrolyte that dissociates into two ions: K⁺ and Cl⁻. Therefore, \( i = 2 \).
- **MgCl₂**: MgCl₂ is also a strong electrolyte that dissociates into three ions: Mg²⁺ and 2 Cl⁻. Therefore, \( i = 3 \).

2. **Calculate the osmotic pressure for each solute:**
- For glucose:
\[ \pi_{\text{glucose}} = 1 \cdot CRT \]
- For KCl:
\[ \pi_{\text{KCl}} = 2 \cdot CRT \]
- For MgCl₂:
\[ \pi_{\text{MgCl₂}} = 3 \cdot CRT \]

3. **Compare the osmotic pressures:**
- Since \( CRT \) is constant for all solutions, we can compare the factors:
- \( \pi_{\text{glucose}} = 1 \cdot CRT \)
- \( \pi_{\text{KCl}} = 2 \cdot CRT \)
- \( \pi_{\text{MgCl₂}} = 3 \cdot CRT \)

4. **Determine the order of osmotic pressure:**
- The order of osmotic pressure from lowest to highest is:
- Glucose (1) < KCl (2) < MgCl₂ (3)

### Final Order:
The osmotic pressure of equimolar solutions of glucose, KCl, and MgCl₂ follows the order:
\[ \text{Glucose} < \text{KCl} < \text{MgCl₂} \]