The osmotic pressure of 6% solution of urea at 258 K is

Explanation

To find the osmotic pressure of a 6% solution of urea at 258 K, we can use the formula for osmotic pressure (π):

\[ \pi = CRT \]

where:
- \( \pi \) = osmotic pressure
- \( C \) = concentration in moles per liter (mol/L)
- \( R \) = gas constant = 0.0821 atm·L/(K·mol)
- \( T \) = temperature in Kelvin

### Step 1: Calculate the number of moles of urea

Given that we have a 6% solution of urea, this means there are 6 grams of urea in 100 mL of solution.

To find the number of moles of urea, we need the molecular weight of urea (NH₂CONH₂), which is approximately 60 g/mol.

\[ \text{Number of moles (n)} = \frac{\text{mass (g)}}{\text{molecular weight (g/mol)}} \]

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

### Step 2: Convert the volume of the solution to liters

Since the volume of the solution is given in mL, we need to convert it to liters:

\[ \text{Volume (V)} = 100 \text{ mL} = 0.1 \text{ L} \]

### Step 3: Calculate the concentration (C)

Now we can calculate the concentration (C) in moles per liter:

\[ C = \frac{n}{V} = \frac{0.1 \text{ moles}}{0.1 \text{ L}} = 1 \text{ mol/L} \]

### Step 4: Substitute values into the osmotic pressure formula

Now we can substitute the values into the osmotic pressure formula:

\[ \pi = CRT \]

Substituting the values we have:

\[ \pi = (1 \text{ mol/L}) \times (0.0821 \text{ atm·L/(K·mol)}) \times (258 \text{ K}) \]

### Step 5: Calculate the osmotic pressure

Now we perform the multiplication:

\[ \pi = 1 \times 0.0821 \times 258 \approx 21.18 \text{ atm} \]

### Final Answer

The osmotic pressure of the 6% solution of urea at 258 K is approximately **21.18 atm**.

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