Question

At what temperature does a 5% (w/v) solution of glucose produce 7 atmospheric osmotic pressure? \([ R = 0.0821 \, \text{L} \cdot \text{atm} / \text{K mol} ]\)

Hint:

In osmotic pressure calculations, ensure that you convert the volume to liters and use the appropriate molar mass for the solute.

Answer 1

Step 1: Use the osmotic pressure formula.
The formula for osmotic pressure is: \[ \Pi = \dfrac{nRT}{V} \] Where: \(\Pi\) = Osmotic pressure = 7 atm \(n\) = Moles of solute \(R\) = Ideal gas constant = 0.0821 L atm / K mol \(T\) = Temperature in Kelvin \(V\) = Volume of the solution = 1 L (assumed for simplicity)
Step 2: Moles of glucose (n).
Given that the solution is 5% (w/v), this means 5 g of glucose is present in 100 ml of solution. So, for 1 liter (1000 ml), there will be: \[ \text{Mass of glucose} = 5 \, \text{g} \times 10 = 50 \, \text{g} \] Molar mass of glucose \(C_6H_{12}O_6\) is: \[ 6(12) + 12(1) + 6(16) = 180 \, \text{g/mol} \] Moles of glucose = \(\dfrac{50}{180} = 0.2778 \, \text{mol}\)
Step 3: Calculate temperature (T).
Substitute the known values into the osmotic pressure equation: \[ 7 = \dfrac{0.2778 \times 0.0821 \times T}{1} \] Solving for \(T\): \[ T = \dfrac{7}{0.2778 \times 0.0821} = 307.7 \, \text{K} \] Final Answer: \[ \boxed{307.7 \, \text{K}} \]