Question

200 ml of an aqueous solution contains 3.6 g of Glucose and 1.2 g of Urea maintained at a temperature equal to 27$^{\circ}$C. What is the Osmotic pressure of the solution in atmosphere units?
Given Data R = 0.082 L atm K$^{-1}$ mol$^{-1}$
Molecular Formula: Glucose = C$_6$H$_{12}$O$_6$, Urea = NH$_2$CONH$_2$

• A. 6.24
• B. 1.56
• C. 9.84
• D. 4.92

Hint:

To solve osmotic pressure problems, always start by calculating the total moles of solute in the solution, then apply the formula $\Pi = \dfrac{nRT}{V}$ where $n$ is the total number of moles, $R$ is the gas constant, $T$ is the temperature in Kelvin, and $V$ is the volume in liters.

Answer 1

The Correct Option is D

Osmotic pressure, $\Pi$, is given by the formula: \[ \Pi = \dfrac{nRT}{V} \] Where: - $n$ = total moles of solute - $R$ = ideal gas constant = 0.082 L atm K$^{-1}$ mol$^{-1}$ - $T$ = temperature in Kelvin = 27$^{\circ}$C = 300 K - $V$ = volume in liters = 0.2 L First, calculate the moles of Glucose and Urea:
1. Moles of Glucose: \[ \text{Molar mass of Glucose} = 6 \times 12 + 12 \times 1 + 6 \times 16 = 180 \, \text{g/mol} \] \[ \text{Moles of Glucose} = \dfrac{3.6 \, \text{g}}{180 \, \text{g/mol}} = 0.02 \, \text{mol} \]
2. Moles of Urea: \[ \text{Molar mass of Urea} = 2 \times 14 + 4 \times 1 + 2 \times 16 = 60 \, \text{g/mol} \] \[ \text{Moles of Urea} = \dfrac{1.2 \, \text{g}}{60 \, \text{g/mol}} = 0.02 \, \text{mol} \] Total moles of solute, $n = 0.02 + 0.02 = 0.04 \, \text{mol}$ Now, using the osmotic pressure formula: \[ \Pi = \dfrac{(0.04 \, \text{mol})(0.082 \, \text{L atm K}^{-1} \text{mol}^{-1})(300 \, \text{K})}{0.2 \, \text{L}} \] \[ \Pi = 4.92 \, \text{atm} \]
Thus, the osmotic pressure of the solution is 4.92 atm.