Question

0.2 M aqueous solution of glucose has osmotic pressure 4.9 atm at 300 K. What is the concentration of glucose if it has osmotic pressure 1.5 atm at same temperature?

• A. 0.03 M
• B. 0.04 M
• C. 0.05 M
• D. 0.06 M

Hint:

Logic Tip: The direct proportionality $\pi_1 / M_1 = \pi_2 / M_2$ is much faster during an exam because it completely bypasses the need to use or remember the exact value of the gas constant $R$.

Answer 1

The Correct Option is D

Concept:
The osmotic pressure ($\pi$) of a dilute solution is directly proportional to its molar concentration ($M$) and the absolute temperature ($T$). It is governed by the van 't Hoff equation: $$\pi = iMRT$$ where $i$ is the van 't Hoff factor, $R$ is the universal gas constant, and $T$ is temperature. Because glucose is a non-electrolyte, $i=1$.
Step 1: Establish the relationship between the two states.
Since both solutions are of the same solute (glucose) at the same temperature (300 K), the values of $i$, $R$, and $T$ are constant. Thus, osmotic pressure is directly proportional to concentration: $\pi \propto M$. We can set up a direct ratio between the two states: $$\frac{\pi_1}{M_1} = \frac{\pi_2}{M_2}$$
Step 2: Substitute the known values and solve for $M_2$.
Given: $\pi_1 = 4.9\text{ atm}$ $M_1 = 0.2\text{ M}$ $\pi_2 = 1.5\text{ atm}$ Substitute these values into the ratio: $$\frac{4.9}{0.2} = \frac{1.5}{M_2}$$ Rearrange to solve for $M_2$: $$M_2 = \frac{1.5 \times 0.2}{4.9} = \frac{0.30}{4.9}$$ $$M_2 \approx 0.0612\text{ M}$$
Step 3: Alternative exact method using the gas constant.
Alternatively, we can calculate $M_2$ directly using the formula $\pi = MRT$ with standard values ($R \approx 0.0821\text{ L atm K}^{-1}\text{mol}^{-1}$): $$M_2 = \frac{\pi_2}{RT}$$ $$M_2 = \frac{1.5}{0.082 \times 300} = \frac{1.5}{24.6} \approx 0.0609\text{ M}$$ Both methods yield a value that rounds to $0.06\text{ M}$, matching option (D).