Question:medium

Calculate the osmotic pressure of 0.2 M aqueous solution of electrolyte at 300 K. If van't Hoff factor is 1.6 [R = 0.0821 atm dm$^3$ K$^{-1}$ mol$^{-1}$].

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Always include the van't Hoff factor ($i$) in colligative property calculations for electrolytes.
Updated On: May 16, 2026
  • 7.21 atm
  • 7.88 atm
  • 8.81 atm
  • 8.32 atm
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: Understanding the Question:
The question asks to calculate the osmotic pressure of an electrolyte solution, given its molarity, temperature, van't Hoff factor, and the universal gas constant.
Step 2: Key Formula or Approach:
The formula for osmotic pressure (\(\Pi\)) of an electrolyte solution is given by the van't Hoff equation:
\[ \Pi = i \times C \times R \times T \] where:
- \(i\) is the van't Hoff factor
- \(C\) is the molar concentration (Molarity) of the solution
- \(R\) is the universal gas constant
- \(T\) is the absolute temperature in Kelvin
Step 3: Detailed Explanation:
Given values:
- Molarity, \(C\) = 0.2 M (or 0.2 mol/dm\(^3\))
- Temperature, \(T\) = 300 K
- van't Hoff factor, \(i\) = 1.6
- Gas constant, \(R\) = 0.0821 atm dm\(^3\) K\(^{-1}\) mol\(^{-1}\)
Substitute these values into the osmotic pressure formula:
\[ \Pi = (1.6) \times (0.2 \text{ mol/dm}^3) \times (0.0821 \text{ atm dm}^3 \text{ K}^{-1} \text{ mol}^{-1}) \times (300 \text{ K}) \] Now, perform the calculation:
\[ \Pi = 1.6 \times 0.2 \times 0.0821 \times 300 \text{ atm} \] \[ \Pi = 0.32 \times 0.0821 \times 300 \text{ atm} \] \[ \Pi = 0.32 \times 24.63 \text{ atm} \] \[ \Pi = 7.8816 \text{ atm} \] Rounding to two decimal places, the osmotic pressure is 7.88 atm.
Step 4: Final Answer:
The calculated osmotic pressure is 7.88 atm, which matches option (B).
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