Arrange the following solutions in increasing order of their osmotic pressures .
(i)34.2 g/litre surcrose
(ii)60 g/litre of urea
(iii)90 g/litre of glucose
(iv)58.5 g/litre of sodium chloride

Options

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

Explanation

To solve the problem of arranging the given solutions in increasing order of their osmotic pressures, we will follow these steps:

### Step 1: Understand the formula for osmotic pressure
The osmotic pressure (π) can be calculated using the formula:
\[ \pi = i \cdot C \cdot R \cdot T \]
Where:
- \( i \) = Van't Hoff factor (number of particles the solute dissociates into)
- \( C \) = Concentration of the solution in moles per liter (Molarity)
- \( R \) = Gas constant (constant for all solutions)
- \( T \) = Temperature (constant for all solutions)

Since \( R \) and \( T \) are constant for all solutions, we can focus on \( i \cdot C \).

### Step 2: Calculate the concentration (C) for each solution
1. **Sucrose**:
- Given: 34.2 g/L
- Molar mass of sucrose = 342 g/mol
- Number of moles = \( \frac{34.2 \text{ g}}{342 \text{ g/mol}} = 0.1 \text{ moles} \)
- Concentration \( C = 0.1 \text{ M} \) (since volume is 1 L)
- Van't Hoff factor \( i = 1 \) (sucrose does not dissociate)
- \( i \cdot C = 1 \cdot 0.1 = 0.1 \)

2. **Urea**:
- Given: 60 g/L
- Molar mass of urea = 60 g/mol
- Number of moles = \( \frac{60 \text{ g}}{60 \text{ g/mol}} = 1 \text{ mole} \)
- Concentration \( C = 1 \text{ M} \)
- Van't Hoff factor \( i = 1 \) (urea does not dissociate)
- \( i \cdot C = 1 \cdot 1 = 1 \)

3. **Glucose**:
- Given: 90 g/L
- Molar mass of glucose = 180 g/mol
- Number of moles = \( \frac{90 \text{ g}}{180 \text{ g/mol}} = 0.5 \text{ moles} \)
- Concentration \( C = 0.5 \text{ M} \)
- Van't Hoff factor \( i = 1 \) (glucose does not dissociate)
- \( i \cdot C = 1 \cdot 0.5 = 0.5 \)

4. **Sodium Chloride (NaCl)**:
- Given: 58.5 g/L
- Molar mass of NaCl = 58.5 g/mol
- Number of moles = \( \frac{58.5 \text{ g}}{58.5 \text{ g/mol}} = 1 \text{ mole} \)
- Concentration \( C = 1 \text{ M} \)
- Van't Hoff factor \( i = 2 \) (NaCl dissociates into Na⁺ and Cl⁻)
- \( i \cdot C = 2 \cdot 1 = 2 \)

### Step 3: Compare \( i \cdot C \) values
- Sucrose: \( i \cdot C = 0.1 \)
- Urea: \( i \cdot C = 1 \)
- Glucose: \( i \cdot C = 0.5 \)
- Sodium Chloride: \( i \cdot C = 2 \)

### Step 4: Arrange in increasing order of osmotic pressure
The increasing order of osmotic pressures based on \( i \cdot C \) values is:
1. Sucrose (0.1)
2. Glucose (0.5)
3. Urea (1)
4. Sodium Chloride (2)

### Final Answer
The solutions in increasing order of their osmotic pressures are:
1. Sucrose (34.2 g/L)
2. Glucose (90 g/L)
3. Urea (60 g/L)
4. Sodium Chloride (58.5 g/L)

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