A solution containing 0.102 g of an unknown compound dissolved in 100. mL of water has an osmotic pressure of 28.1 mmHg at 20.°C. What is the molar mass of the compound? (R = 0.08206 L • atm/K • mol, 1 atm = 760 mmHg) A) 663 g/mol B) 0.872 g/mol C) 1.15 g/mol D) 727 g/mol E) 1.10 × 10² g/mol

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To find the molar mass of the compound, use the formula for osmotic pressure: $\pi = \frac{n}{V}RT$, where $\pi$ is the osmotic pressure, $n$ is the number of moles of solute, $V$ is the volume of the solution in liters, $R$ is the ideal gas constant, and $T$ is the temperature in Kelvin. First, convert the osmotic pressure to atm: $28.1 \, \text{mmHg} \times \frac{1 \, \text{atm}}{760 \, \text{mmHg}} = 0.037 \, \text{atm}$. Convert the temperature to Kelvin: $20°C + 273.15 = 293.15 \, K$. The volume is already in liters. Solve for $n$: $n = \frac{\pi V}{RT} = \frac{0.037 \, \text{atm} \times 0.1 \, \text{L}}{0.08206 \, \text{L atm K}^{-1} \text{mol}^{-1} \times 293.15 \, K} = 1.54 \times 10^{-4} \, \text{mol}$. Finally, calculate the molar mass: $\frac{0.102 \, \text{g}}{1.54 \times 10^{-4} \, \text{mol}} = 663 \, \text{g/mol}$.

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