Osmotic pressure of solution containing 0.6 g urea and 3.42 g sugar in 100 ml at `27^@C`

Options

A. 492 atm
B. 4.92 atm
C. 49.2 atm
D. 28.1 atm

Answer

To find the osmotic pressure of a solution containing 0.6 g of urea and 3.42 g of sugar in 100 ml at 27°C, we will follow these steps:

### Step 1: Calculate the molarity of urea
1. **Determine the molecular weight of urea (NH2CONH2)**:
- Nitrogen (N) = 14 g/mol (2 atoms)
- Carbon (C) = 12 g/mol (1 atom)
- Oxygen (O) = 16 g/mol (1 atom)
- Hydrogen (H) = 1 g/mol (4 atoms)
- Molecular weight of urea = (2 × 14) + (1 × 12) + (1 × 16) = 28 + 12 + 16 = 60 g/mol

2. **Calculate the number of moles of urea**:
\[ \text{Moles of urea} = \frac{\text{mass}}{\text{molecular weight}} = \frac{0.6 \text{ g}}{60 \text{ g/mol}} = 0.01 \text{ mol} \]

3. **Convert the volume from ml to liters**:
\[ \text{Volume in liters} = \frac{100 \text{ ml}}{1000} = 0.1 \text{ L} \]

4. **Calculate the molarity of urea**:
\[ \text{Molarity (C)} = \frac{\text{moles}}{\text{volume in L}} = \frac{0.01 \text{ mol}}{0.1 \text{ L}} = 0.1 \text{ M} \]

### Step 2: Calculate the molarity of sugar
1. **Determine the molecular weight of sugar (C12H22O11)**:
- Carbon (C) = 12 g/mol (12 atoms)
- Hydrogen (H) = 1 g/mol (22 atoms)
- Oxygen (O) = 16 g/mol (11 atoms)
- Molecular weight of sugar = (12 × 12) + (22 × 1) + (11 × 16) = 144 + 22 + 176 = 342 g/mol

2. **Calculate the number of moles of sugar**:
\[ \text{Moles of sugar} = \frac{3.42 \text{ g}}{342 \text{ g/mol}} = 0.01 \text{ mol} \]

3. **Calculate the molarity of sugar**:
\[ \text{Molarity of sugar} = \frac{0.01 \text{ mol}}{0.1 \text{ L}} = 0.1 \text{ M} \]

### Step 3: Calculate the total molarity of the solution
\[ \text{Total molarity} = \text{Molarity of urea} + \text{Molarity of sugar} = 0.1 \text{ M} + 0.1 \text{ M} = 0.2 \text{ M} \]

### Step 4: Calculate the osmotic pressure using the formula
The formula for osmotic pressure (\(\pi\)) is given by:
\[ \pi = CRT \]
where:
- \(C\) = total molarity (0.2 M)
- \(R\) = gas constant = 0.0821 L·atm/(K·mol)
- \(T\) = temperature in Kelvin = 27°C + 273 = 300 K

Substituting the values:
\[ \pi = 0.2 \text{ M} \times 0.0821 \frac{\text{L·atm}}{\text{K·mol}} \times 300 \text{ K} \]
\[ \pi = 4.926 \text{ atm} \approx 4.92 \text{ atm} \]

### Final Answer
The osmotic pressure of the solution is approximately **4.92 atm**.

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