Question

Calculate the osmotic pressure of 0.03 mole of non electrolyte solute dissolved in $0.1\text{dm}^3$ of water at $300 \text{ K}$. $[\text{R} = 0.082\text{dm}^3 \text{ atm mol}^{-1} \text{ K}^{-1}]$}

• A. 7.4 atm
• B. 6.4 atm
• C. 8.0 atm
• D. 5.6 atm

Hint:

Shortcut: $R \times 300$ is approximately 24.6. Just multiply your molarity ($n/V$) by 24.6 for quick results at 300K.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Osmotic pressure ($\pi$) is an important colligative property of solutions that depends strictly on the concentration of solute particles, not their chemical identity. It can be calculated using an equation mathematically identical to the ideal gas law.
Step 2: Key Formula or Approach:
The standard formula for calculating osmotic pressure is: \[ \pi = iCRT \quad \text{or} \quad \pi = i\frac{n}{V}RT \] where: $i$ = van 't Hoff factor (which is $1$ for a non-electrolyte) $n$ = number of moles of solute $V$ = volume of solution in liters ($\text{dm}^3$) $R$ = ideal gas constant $T$ = absolute temperature in Kelvin
Step 3: Detailed Explanation:
Let's clearly list the numerical values given in the problem statement: Number of moles ($n$) = $0.03 \text{ mol}$ Volume of solution ($V$) = $0.1 \text{ dm}^3$ (Note: Since it is a dilute solution, the volume of the solvent water is an acceptable approximation for the volume of the solution). Temperature ($T$) = $300 \text{ K}$ Gas constant ($R$) = $0.082 \text{ dm}^3 \text{ atm mol}^{-1} \text{ K}^{-1}$ Because the solute is explicitly described as a "non-electrolyte", it does not dissociate into ions in solution, meaning $i = 1$. Substitute these values directly into the osmotic pressure formula: \[ \pi = (1) \times \left( \frac{0.03 \text{ mol}}{0.1 \text{ dm}^3} \right) \times (0.082 \text{ dm}^3 \text{ atm mol}^{-1} \text{ K}^{-1}) \times (300 \text{ K}) \] Simplify the concentration term first: \[ \pi = (0.3 \text{ mol dm}^{-3}) \times (0.082 \text{ dm}^3 \text{ atm mol}^{-1} \text{ K}^{-1}) \times (300 \text{ K}) \] Perform the multiplication: \[ \pi = 0.3 \times 0.082 \times 300 \text{ atm} \] \[ \pi = 0.3 \times 24.6 \text{ atm} \] \[ \pi = 7.38 \text{ atm} \] Rounding this result to one decimal place to appropriately match the provided options gives $7.4 \text{ atm}$.
Step 4: Final Answer:
The calculated osmotic pressure is 7.4 atm.