Question

A solution of 8 g of a certain organic compound in 2 dm³ water develops osmotic pressure 0.6 atm at 300 K. Calculate the molar mass of compound. [R = 0.082 atm dm^3 K^{-1} mol^{-1}]

• A. $148\ \text{g mol}^{-1}$
• B. $164\ \text{g mol}^{-1}$
• C. $172\ \text{g mol}^{-1}$
• D. $180\ \text{g mol}^{-1}$

Hint:

Key Exam Tip:
The van't Hoff equation ($\pi = CRT$) is fundamental for osmotic pressure calculations. Ensure that volume is in liters or dm³ and temperature is in Kelvin. Molar mass is calculated using $M = \frac{mRT}{\pi V}$.

Answer 1

The Correct Option is B

The osmotic pressure ($\pi$) of a solution is related to its molar concentration ($C$), the ideal gas constant ($R$), and the absolute temperature ($T$) by the van't Hoff equation: $\pi = CRT$.
The molar concentration ($C$) is defined as the number of moles of solute ($n$) divided by the volume of the solution ($V$): $C = n/V$.
The number of moles ($n$) can be expressed as the mass of the solute ($m$) divided by its molar mass ($M$): $n = m/M$.
Substituting these into the van't Hoff equation, we get:
$\pi = \left(\frac{m}{M}\right) \frac{R T}{V}$
We are given:
Mass of the organic compound ($m$) = 8 g
Volume of the solution ($V$) = 2 dm³
Osmotic pressure ($\pi$) = 0.6 atm
Temperature ($T$) = 300 K
Ideal gas constant ($R$) = 0.082 atm dm³ K⁻¹ mol⁻¹
We need to calculate the molar mass ($M$). Rearranging the equation to solve for $M$:
$M = \frac{m \times R \times T}{\pi \times V}$
Now, substitute the given values into the formula:
$M = \frac{(8 \text{ g}) \times (0.082 \text{ atm dm³ K⁻¹ mol⁻¹}) \times (300 \text{ K})}{(0.6 \text{ atm}) \times (2 \text{ dm³})}$
First, calculate the numerator:
$8 \times 0.082 \times 300 = 196.8$
Next, calculate the denominator:
$0.6 \times 2 = 1.2$
Finally, calculate the molar mass:
$M = \frac{196.8}{1.2} \text{ g mol⁻¹}$
$M = 164 \text{ g mol⁻¹}$
Final Answer: \(\boxed{B}\)