Question

At 300 K, 6 g of urea was dissolved in 500 mL of water. What is the osmotic pressure (in atm) of the resultant solution? (R = 0.082 L atm K$^{-1}$ mol$^{-1}$) (C=12;N=14;O=16;H=1)

• A. 0.492
• B. 2.46
• C. 4.92
• D. 49.2

Hint:

Remember that osmotic pressure is directly proportional to the number of moles of solute, the temperature, and inversely proportional to the volume of the solution.

Answer 1

The Correct Option is C

The osmotic pressure \(\pi\) is given by the formula: \[ \pi = \frac{nRT}{V} \] Where: - \(n\) = number of moles of urea,
- \(R\) = gas constant = 0.082 L atm K$^{-1}$ mol$^{-1}$,
- \(T\) = temperature in Kelvin = 300 K,
- \(V\) = volume of solution in liters = 500 mL = 0.5 L.
First, calculate the number of moles of urea: \[ \text{Molar mass of urea (NH}_2\text{CONH}_2) = 12 + 2(1) + 16 + 2(14) = 60 \, \text{g/mol}. \] \[ n = \frac{\text{mass of urea}}{\text{molar mass}} = \frac{6}{60} = 0.1 \, \text{mol}. \] Now, calculate the osmotic pressure: \[ \pi = \frac{0.1 \times 0.082 \times 300}{0.5} = 4.92 \, \text{atm}. \]