Question

A solution of 8 g of 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 dm³ K⁻¹ mol⁻¹]

• A. 148 g mol⁻¹
• B. 164 g mol⁻¹
• C. 172 g mol⁻¹
• D. 180 g mol⁻¹

Answer 1

Answer: (B)

164 g mol⁻¹

Answer 2

1. Calculate the number of moles of the compound:
   $\text{Moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{8}{M}$

2. Determine the molarity of the solution:
   $\text{Molarity}, c = \frac{(8/M)}{2} = \frac{4}{M}$

3. Use the osmotic pressure formula:
   $\pi = cRT$
   Substitute the values:
   $0.6 = \left(\frac{4}{M}\right)(0.082)(300)$

4. Solve for $M$:
   $4 \times 0.082 \times 300 = 98.4$
   So, $0.6 = \frac{98.4}{M}$
   $\Rightarrow M = \frac{98.4}{0.6} = 164 \, \text{g/mol}$