Step 1: Understanding the Question:
The problem asks for the calculation of osmotic pressure (\(\pi\)) of a glucose solution, given its molar concentration (\(C\)), temperature (\(T\)), and the ideal gas constant (\(R\)).
Step 2: Key Formula or Approach:
Osmotic pressure is a colligative property calculated using the van 't Hoff equation:
\[
\pi = iCRT
\]
where:
- \(\pi\) is the osmotic pressure.
- \(i\) is the van 't Hoff factor (number of particles the solute dissociates into).
- \(C\) is the molar concentration.
- \(R\) is the ideal gas constant.
- \(T\) is the absolute temperature in Kelvin.
Step 3: Detailed Explanation:
(i) Identify the given values and convert units:
- Molar concentration, \(C = 0.1\,M\) (or \(0.1\,mol/L\)).
- Gas constant, \(R = 0.0821\,L\,atm\,mol^{-1}K^{-1}\).
- Temperature, \(T = 27^\circ C\). We must convert this to Kelvin: \(T(K) = T(^\circ C) + 273 = 27 + 273 = 300\,K\).
- Van 't Hoff factor, \(i\): Glucose (\(C_6H_{12}O_6\)) is a non-electrolyte, meaning it does not dissociate in solution. Therefore, \(i = 1\).
(ii) Substitute the values into the formula:
The formula simplifies to \(\pi = CRT\) since \(i=1\).
\[
\pi = (0.1 \, \text{mol/L}) \times (0.0821 \, \text{L atm mol}^{-1}K^{-1}) \times (300 \, \text{K})
\]
(iii) Calculate the osmotic pressure:
\[
\pi = 0.1 \times 0.0821 \times 300
\]
\[
\pi = 0.0821 \times 30
\]
\[
\pi = 2.463 \, \text{atm}
\]
Step 4: Final Answer:
The osmotic pressure of the solution is approximately \(2.46\,\text{atm}\).