Question

The osmotic pressure of \( 0.01\,M \) aqueous solution of urea at \( 300\,K \) is \((R = 0.082\,\text{lit atm mol}^{-1}\text{K}^{-1})\)

• A. \( 0.0082\,\text{atm} \)
• B. \( 0.082\,\text{atm} \)
• C. \( 2.46\,\text{atm} \)
• D. \( 24.6\,\text{atm} \)
• E. \( 0.246\,\text{atm} \)

Hint:

For non-electrolytes like urea, always take \( i=1 \). Use \( \pi = MRT \) directly.

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
Osmotic pressure is a colligative property of solutions, which depends on the concentration of solute particles, not their identity. The question asks to calculate the osmotic pressure for a urea solution of a given concentration and temperature.
Step 2: Key Formula or Approach:
The formula for osmotic pressure (\(\Pi\)) is given by the van't Hoff equation:
\[ \Pi = i \cdot C \cdot R \cdot T \] where:
\(\Pi\) = Osmotic pressure
\(i\) = van't Hoff factor (number of particles the solute dissociates into)
\(C\) = Molar concentration of the solution
\(R\) = Ideal gas constant
\(T\) = Absolute temperature in Kelvin
Step 3: Detailed Explanation:
First, identify the values for each variable in the formula.

van't Hoff factor (\(i\)): Urea (NH\(_2\)CONH\(_2\)) is a non-electrolyte, meaning it does not dissociate or associate in water. Therefore, its van't Hoff factor is \(i = 1\).

Concentration (\(C\)): Given as 0.01 M.

Gas constant (\(R\)): Given as 0.082 lit atm mol\(^{-1}\) K\(^{-1}\).

Temperature (\(T\)): Given as 300 K.

Now, substitute these values into the osmotic pressure equation:
\[ \Pi = (1) \times (0.01 \text{ mol/lit}) \times (0.082 \text{ lit atm mol}^{-1} \text{ K}^{-1}) \times (300 \text{ K}) \] Let's perform the calculation:
\[ \Pi = 0.01 \times (0.082 \times 300) \] \[ \Pi = 0.01 \times 24.6 \] \[ \Pi = 0.246 \text{ atm} \] Step 4: Final Answer:
The osmotic pressure of the 0.01 M urea solution at 300 K is 0.246 atm. This corresponds to option (E).