For the equimolal solutions, increasing order of osmotic pressure is

1. 1 m \(C_6H_5COOH\)
2. 1 m \(BaCl_2\)
3. 1 m \(KCl\)
4. 1 m glucose
5. 1 m \(H_3PO_4\)

Choose the correct answer from the options given below:

D < A < C < B <  E

The correct answer is option 4. D < A < C < B <  E.

To determine the increasing order of osmotic pressure for the given solutions, we need to carefully understand the role of the van 't Hoff factor (\(i\)) and how it affects osmotic pressure. Osmotic pressure is given by the formula:

\(\pi = i \cdot M \cdot R \cdot T\)

Where:

\(\pi\) is the osmotic pressure,

\(i\) is the van 't Hoff factor (the number of particles the solute dissociates into),

\(M\) is the molarity of the solution (in this case, 1 molal for each solution),

\(R\) is the gas constant (a constant value),

\(T\) is the temperature (constant for this problem).

The osmotic pressure is directly proportional to the van 't Hoff factor \(i\). The higher the \(i\), the greater the osmotic pressure.

In the given problem, since the molality is 1 for all solutions and the temperature is assumed to be constant, the only variable affecting osmotic pressure is the van 't Hoff factor \(i\).

Analyzing the Given Compounds:

A. \(C_6H_5COOH\) (Benzoic acid): 

Benzoic acid is a weak acid, meaning it does not dissociate completely in water. 

When it dissolves in water, it is partially dissociated as \(C_6H_5COO^- + H^+\)molecules, so \(i = 2\).

But due to the partial dissociation it will be less than 2.
   
B. \(BaCl_2\) (Barium chloride):

Barium chloride is an ionic compound that dissociates completely in water.

It dissociates into three ions: Ba\(^{2+}\) and 2 Cl\(^-\).

Therefore, \(i = 3\), because it produces 3 particles per formula unit.

C. \(KCl\) (Potassium chloride):

Potassium chloride is also an ionic compound and dissociates completely in water.

It dissociates into two ions: K\(^+\) and Cl\(^-\).

Therefore, \(i = 2\), since it produces 2 particles per formula unit.

D. Glucose \((C_6H_{12}O_6)\):

Glucose is a molecular compound and does not dissociate in solution.

Therefore, \(i = 1\), because it remains as intact glucose molecules in solution.

E. \(H_3PO_4\) (Phosphoric acid):

Phosphoric acid is a triprotic acid, meaning it can dissociate in multiple steps:

First, it dissociates into \(H^+\) and \(H_2PO_4^-\),

Then \(H_2PO_4^-\) dissociates into \(H^+\) and \(HPO_4^{2-}\),

And \(HPO_4^{2-}\) dissociates into \(H^+\) and \(PO_4^{3-}\).

Therefore, \(H_3PO_4\) dissociates into a total of 4 particles: 3 \(H^+\) ions and 1 \(PO_4^{3-}\) ion (if all dissociation steps are complete).

So, \(i = 4\).

Increasing Order of Osmotic Pressure:

Since osmotic pressure is directly proportional to the van 't Hoff factor, the solution with the smallest \(i\) will have the lowest osmotic pressure, and the solution with the largest \(i\) will have the highest osmotic pressure.

Thus, the increasing order of osmotic pressure is:

\(\text{D (Glucose)} = 1, \text{A (Benzoic acid)} < 2, \text{C (KCl)} = 2, \text{B (BaCl2)} = 3, \text{E (H3PO4)} = 4\)

This corresponds to Option 4: D < A < C < B < E