The osmotic pressure of 6% solution of urea at 300 K is

Answer

To calculate the osmotic pressure of a 6% solution of urea at 300 K, we can follow these steps:

### Step 1: Understand the formula for osmotic pressure
The osmotic pressure (π) can be calculated using the formula:
\[ \pi = cRT \]
where:
- \(c\) = concentration of the solution in moles per liter (mol/L)
- \(R\) = universal gas constant = 0.0821 L·atm/(K·mol)
- \(T\) = temperature in Kelvin (K)

### Step 2: Determine the concentration of urea
A 6% solution means there are 6 grams of urea in 100 mL of solution. First, we need to convert this to moles.

1. **Calculate the number of moles of urea**:
\[ \text{Number of moles (n)} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]
The molar mass of urea (NH₂CONH₂) is approximately 60 g/mol.

\[ n = \frac{6 \text{ g}}{60 \text{ g/mol}} = 0.1 \text{ mol} \]

2. **Convert the volume from mL to L**:
\[ \text{Volume (V)} = 100 \text{ mL} = 0.1 \text{ L} \]

3. **Calculate the concentration (c)**:
\[ c = \frac{n}{V} = \frac{0.1 \text{ mol}}{0.1 \text{ L}} = 1 \text{ mol/L} \]

### Step 3: Substitute values into the osmotic pressure formula
Now we can substitute the values into the osmotic pressure formula:
\[ \pi = cRT \]
Substituting the known values:
- \(c = 1 \text{ mol/L}\)
- \(R = 0.0821 \text{ L·atm/(K·mol)}\)
- \(T = 300 \text{ K}\)

\[ \pi = 1 \text{ mol/L} \times 0.0821 \text{ L·atm/(K·mol)} \times 300 \text{ K} \]

### Step 4: Calculate the osmotic pressure
\[ \pi = 0.0821 \times 300 = 24.63 \text{ atm} \]

### Final Answer
The osmotic pressure of the 6% solution of urea at 300 K is approximately **24.63 atm**.

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