Question

The osmotic pressure (in atm) of an aqueous solution containing 0.01 mol of NaCl (degree of dissociation 0.94) and 0.03 mol of glucose in 500 mL at \(27^\circ\text{C}\) is \((R = 0.082\text{L atm K}^{-1}\text{mol}^{-1})\)

• A. 2.43
• B. 4.23
• C. 3.24
• D. 3.42

Hint:

Osmotic pressure is a colligative property. When multiple solutes are present, the pressures exerted by each are additive. \(\pi_{\text{mix}} = \pi_1 + \pi_2 + \dots = (\sum i_j C_j)RT\).

Answer 1

The Correct Option is A

Step 1: Calculate Van't Hoff Factor for Electrolytes:
For NaCl (Electrolyte): \(\text{NaCl} \rightarrow \text{Na}^+ + \text{Cl}^-\) (\(n=2\)). Given \(\alpha = 0.94\). \(i_{\text{NaCl}} = 1 + (n-1)\alpha = 1 + (2-1)(0.94) = 1.94\). For Glucose (Non-electrolyte): \(i_{\text{glu}} = 1\).
Step 2: Calculate Effective Molarity (C):
Volume \(V = 500 \text{ mL} = 0.5 \text{ L}\). Temperature \(T = 27^\circ\text{C} = 300 \text{ K}\). Total effective moles (\(n_{\text{total}}\)): \(n_{\text{eff}} = (i_{\text{NaCl}} \times n_{\text{NaCl}}) + (i_{\text{glu}} \times n_{\text{glu}})\) \(n_{\text{eff}} = (1.94 \times 0.01) + (1 \times 0.03)\) \(n_{\text{eff}} = 0.0194 + 0.03 = 0.0494 \text{ mol}\).
Step 3: Calculate Osmotic Pressure (\(\pi\)):
\[ \pi = \frac{n_{\text{eff}}}{V} \times R \times T \] \[ \pi = \frac{0.0494}{0.5} \times 0.082 \times 300 \] \[ \pi = 0.0988 \times 24.6 \] \[ \pi \approx 2.43 \text{ atm} \] Final Answer:
2.43 atm.