The osmotic pressure of 1.00 g of a polymer dissolved in benzene to give 200 mL of solution is 1.50 kPa at 25^$\omicron$C .Estimate the average molar mass of the polymer.The gas law constant is 0.0821 L . atm . mol ^- ¹ . K ^- ¹.

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The osmotic pressure ($\Pi$) can be calculated using the formula $\Pi = \frac{n}{V}RT$, where $n$ is the number of moles of solute, $V$ is the volume of solution in liters, $R$ is the gas constant, and $T$ is the temperature in Kelvin. Given that $R = 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1}$, $T = 298 \, \text{K}$ (25°C + 273), and $\Pi = 1.50 \, \text{kPa} = 1.50/101.3 \, \text{atm}$ (since 1 atm = 101.3 kPa), and $V = 0.200 \, \text{L}$, we can solve for $n$, and then use the mass of the polymer to find its molar mass. First, convert the osmotic pressure to atm: $1.50 \, \text{kPa} = 1.50/101.3 \approx 0.0148 \, \text{atm}$.Then, solve for $n$: $0.0148 = \frac{n}{0.200} \times 0.0821 \times 298$.Solving for $n$ gives approximately $0.000121 \, \text{mol}$.Finally, the molar mass $M$ is the mass of the polymer divided by the number of moles: $M = \frac{1.00 \, \text{g}}{0.000121 \, \text{mol}} \approx 8260 \, \text{g/mol}$.

The Osmotic Pressure of 1 ο\omicronο C Estimate the Average Molar Mass of the Polymer

The Osmotic Pressure of 1 $\omicron$ C Estimate the Average Molar Mass of the Polymer