Question

18 g of glucose (molar mass = 180 g/mol) is dissolved in water to prepare 500 ml solution at \(15^{\circ}\text{C}\). Calculate the osmotic pressure of the solution. [\(R = 0.0821 \text{L atm K}^{-1} \text{mol}^{-1}\)]

• A. \(1.65 \text{atm}\)
• B. \(4.73 \text{atm}\)
• C. \(5.57 \text{atm}\)
• D. \(2.34 \text{atm}\)

Hint:

Use pi = CRT with temperature in kelvin.

Answer 1

The Correct Option is B

Step 1: Find the molarity
Moles of glucose \(= \frac{18}{180} = 0.1\ \text{mol}\). Volume is \(500\ \text{mL} = 0.5\ \text{L}\).
\(C = \frac{0.1}{0.5} = 0.2\ \text{M}\).

Step 2: Convert temperature
\(T = 15 + 273 = 288\ \text{K}\).

Step 3: Apply the formula
\[ \pi = CRT = 0.2 \times 0.0821 \times 288 = 4.73\ \text{atm} \]

Step 4: Check the options
Using \(T = 15\) K would give a very small answer, and \(298\) K would give \(4.89\) atm. Neither is listed, so \(4.73\) atm is correct.

Final Answer:
The osmotic pressure is 4.73 atm. \[ \boxed{\text{(B)}\ 4.73\ \text{atm}} \]