6g urea and 1.8g glucose are dissolved in 100 ml `H_(2)O`. What is the osmotic pressure at 300K.

Options

A. 27.06 atm
B. 30 atm
C. 45 atm
D. 13.07 atm

Answer

To find the osmotic pressure of a solution containing 6 g of urea and 1.8 g of glucose dissolved in 100 mL of water at 300 K, we can follow these steps:

### Step 1: Calculate the number of moles of urea and glucose.

**Urea:**
- Molar mass of urea (NH₂CONH₂) = 14 (N) + 2*1 (H) + 12 (C) + 16 (O) = 60 g/mol
- Number of moles of urea = mass / molar mass = 6 g / 60 g/mol = 0.1 moles

**Glucose:**
- Molar mass of glucose (C₆H₁₂O₆) = 6*12 (C) + 12*1 (H) + 6*16 (O) = 180 g/mol
- Number of moles of glucose = mass / molar mass = 1.8 g / 180 g/mol = 0.01 moles

### Step 2: Calculate the total number of moles of solute.

Total moles of solute = moles of urea + moles of glucose
= 0.1 moles + 0.01 moles = 0.11 moles

### Step 3: Calculate the concentration of the solution in moles per liter.

Volume of solution = 100 mL = 0.1 L

Concentration (C) = Total moles of solute / Volume in liters
= 0.11 moles / 0.1 L = 1.1 moles/L

### Step 4: Use the formula for osmotic pressure.

Osmotic pressure (π) is given by the formula:
\[ \pi = C \cdot R \cdot T \]
where:
- \( C \) = concentration in moles/L
- \( R \) = universal gas constant = 0.0821 L·atm/(K·mol)
- \( T \) = temperature in Kelvin

Substituting the values:
\[ \pi = 1.1 \, \text{mol/L} \cdot 0.0821 \, \text{L·atm/(K·mol)} \cdot 300 \, \text{K} \]

### Step 5: Calculate the osmotic pressure.

\[ \pi = 1.1 \cdot 0.0821 \cdot 300 \]
\[ \pi = 27.06 \, \text{atm} \]

Thus, the osmotic pressure of the solution is **27.06 atm**.

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