A dilute aqueous solution of an organic compound soluble in water is formed by dissolving 2.18 grams of the compound in water to form 0.250 L of solution. The resulting solution has an osmotic pressure of 0.638 atm at 25 ° C. What is the molar mass of this organic compound?

Question

Quizplus · Accepted Answer

The molar mass of the organic compound can be calculated using 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 (0.0821 L·atm/(K·mol)), and $T$ is the temperature in Kelvin. Given that $\pi = 0.638$ atm, $V = 0.250$ L, and $T = 25 °C = 298 K$, we can solve for $n$, the number of moles of the solute. Then, the molar mass (M) is the mass of the solute divided by the number of moles ($M = \frac{mass}{n}$).First, convert the temperature to Kelvin: $T = 25 °C + 273 = 298 K$.Rearrange the osmotic pressure formula to solve for $n$: $n = \pi V / RT$.Substitute the given values: $n = (0.638 \, \text{atm} \times 0.250 \, \text{L}) / (0.0821 \, \text{L·atm/(K·mol)} \times 298 \, \text{K})$.Calculate $n$: $n ≈ 0.00654 \, \text{mol}$.Finally, calculate the molar mass: $M = \frac{2.18 \, \text{g}}{0.00654 \, \text{mol}} ≈ 334 \, \text{g/mol}$.

A Dilute Aqueous Solution of an Organic Compound Soluble in Water