Question

What is the osmotic pressure (in atm) of 0.02M aqueous glucose solution at 300 K? (R=0.082 L atm mol\(^{-1} \)K\(^{-1} \))

• A. \( \frac{1}{0.492} \)
• B. \( 0.492 \)
• C. \( 0.988 \)
• D. \( \frac{1}{0.988} \)

Hint:

Use the van't Hoff equation \( \pi = iCRT \) to calculate osmotic pressure. Remember to use the correct value of the van't Hoff factor \( i \) (which is 1 for non-electrolytes like glucose) and ensure that the units of concentration, gas constant, and temperature are consistent to obtain the osmotic pressure in the desired units (atm in this case).

Answer 1

The Correct Option is B

The osmotic pressure \( \pi \) of a solution is given by the van't Hoff equation: $$ \pi = iCRT $$ where: \( i \) is the van't Hoff factor (for glucose, a non-electrolyte, \( i = 1 \)) \( C \) is the molar concentration of the solution \( R \) is the ideal gas constant \( T \) is the absolute temperature Given: Concentration \( C = 0.
02 \) M (mol/L) Temperature \( T = 300 \) K Ideal gas constant \( R = 0.
082 \) L atm mol\(^{-1} \)K\(^{-1} \) Van't Hoff factor for glucose \( i = 1 \) Substitute these values into the van't Hoff equation: $$ \pi = (1) \times (0.
02 \text{ mol/L}) \times (0.
082 \text{ L atm mol}^{-1}\text{K}^{-1}) \times (300 \text{ K}) $$ $$ \pi = 0.
02 \times 0.
082 \times 300 \text{ atm} $$ $$ \pi = 0.
00164 \times 300 \text{ atm} $$ $$ \pi = 0.
492 \text{ atm} $$ The osmotic pressure of the 0.
02M aqueous glucose solution at 300 K is 0.
492 atm.