The osmotic pressure of 5% solution of urea at 273 K is

Options

A. 18.40 atm
B. 18.61 atm
C. 18.59 atm
D. 18.86 atm

Explanation

To find the osmotic pressure of a 5% solution of urea at 273 K, we can follow these steps:

### Step 1: Understand the given information
We have a 5% solution of urea, which means there are 5 grams of urea in 100 grams of solution. The temperature is given as 273 K.

### Step 2: Calculate the number of moles of urea
To find the number of moles of urea, we use the formula:
\[ \text{Number of moles} = \frac{\text{Given mass}}{\text{Molar mass}} \]
The molar mass of urea (NH₂CONH₂) is approximately 60 g/mol. Therefore:
\[ \text{Number of moles of urea} = \frac{5 \text{ g}}{60 \text{ g/mol}} = \frac{1}{12} \text{ mol} \approx 0.0833 \text{ mol} \]

### Step 3: Convert the volume of the solution to liters
Since the solution is 5% w/w, we can assume that the total mass of the solution is 100 g. The density of the solution is approximately 1 g/mL, so:
\[ \text{Volume of solution} = 100 \text{ g} \approx 100 \text{ mL} = 0.1 \text{ L} \]

### Step 4: Use the osmotic pressure formula
The osmotic pressure (\(\pi\)) can be calculated using the formula:
\[ \pi = CRT \]
Where:
- \(C\) is the concentration in moles per liter (mol/L)
- \(R\) is the ideal gas constant (0.0821 L·atm/(K·mol))
- \(T\) is the temperature in Kelvin (K)

First, we calculate the concentration \(C\):
\[ C = \frac{\text{Number of moles}}{\text{Volume in liters}} = \frac{0.0833 \text{ mol}}{0.1 \text{ L}} = 0.833 \text{ mol/L} \]

### Step 5: Calculate the osmotic pressure
Now, substituting the values into the osmotic pressure formula:
\[ \pi = (0.833 \text{ mol/L}) \times (0.0821 \text{ L·atm/(K·mol)}) \times (273 \text{ K}) \]
Calculating this gives:
\[ \pi = 0.833 \times 0.0821 \times 273 \approx 18.67 \text{ atm} \]

### Final Answer
The osmotic pressure of the 5% solution of urea at 273 K is approximately **18.67 atm**.

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