Question

Calculate the concentration of an aqueous solution of non electrolyte at $300\text{ K}$ if its osmotic pressure is $12\text{ atm}$.
$[ \text{R} = 0.0821\text{ atm dm}^3\text{ K}^{-1}\text{ mol}^{-1} ]$

• A. $0.371\text{ M}$
• B. $0.615\text{ M}$
• C. $0.487\text{ M}$
• D. $0.726\text{ M}$

Hint:

Use $T$ in Kelvin always! (K = $^\circ$C + 273.15).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Osmotic pressure ($\pi$) of a solution is directly proportional to its molar concentration ($\text{M}$ or $\text{C}$).
Step 2: Key Formula or Approach:
$\pi = \text{CRT}$
Step 3: Detailed Explanation:
Given:
$\pi = 12\text{ atm}$
$\text{T} = 300\text{ K}$
$\text{R} = 0.0821\text{ atm L mol}^{-1}\text{ K}^{-1}$ (Note: $1\text{ dm}^3 = 1\text{ L}$).
Using the formula $\pi = \text{CRT}$:
\[ 12 = \text{C} \times 0.0821 \times 300 \]
\[ 12 = \text{C} \times 24.63 \]
\[ \text{C} = \frac{12}{24.63} \approx 0.4872\text{ M} \]
Step 4: Final Answer:
The concentration is $0.487\text{ M}$.