Question

Calculate the osmotic pressure of solution containing 0.822 gm of sucrose in 300 mL of water at 298 K. [ Given : Molar mass of sucrose 342 g/mol, R = 0.08205 $dm^3$ atm $K^{-1}$ $mol^{-1}$ ]

Hint:

Always ensure the units of Volume ($V$) match the units of the Gas Constant ($R$). If $R$ is in $dm^3$, $V$ must be in Liters.

Answer 1

Step 1: Understanding the Concept:
Osmotic pressure (\( \pi \)) is a colligative property that depends on the molar concentration of the solute in the solution.
Step 2: Key Formula or Approach:
The formula for osmotic pressure is:
\[ \pi = \frac{W_2 R T}{M_2 V} \]
Where:
\( W_2 \) = mass of solute (0.822 g)
\( R \) = gas constant (0.08205 \( \text{dm}^3 \cdot \text{atm} \cdot \text{K}^{-1} \cdot \text{mol}^{-1} \))
\( T \) = temperature (298 K)
\( M_2 \) = molar mass of solute (342 g/mol)
\( V \) = volume of solution in Liters (\( \text{dm}^3 \))
Step 3: Detailed Explanation:
First, convert volume to liters:
\[ V = 300 \text{ mL} = 0.3 \text{ L} = 0.3 \text{ dm}^3 \]
Now, substitute the values into the formula:
\[ \pi = \frac{0.822 \times 0.08205 \times 298}{342 \times 0.3} \]
\[ \pi = \frac{20.0988}{102.6} \]
\[ \pi \approx 0.19589 \text{ atm} \]
Rounding to three decimal places: \( \pi = 0.196 \text{ atm} \).
Step 4: Final Answer:
The osmotic pressure of the solution is 0.196 atm.