Question

The osmotic pressure of a solution formed by dissolving 75.0 mg of aspirin (c9h8o4) in 0.250 l of water at 25 °c is ________ atm.

Answer 1

To find the osmotic pressure of a solution, calculate the molarity of the aspirin dissolved in water and apply the formula for osmotic pressure incorporating the ideal gas constant and the temperature in Kelvin.The osmotic pressure of a 0.333 M solution of C6H12O6 at 25°C is 8.15 atm.

Step-by-step explanation:

The osmotic pressure of a solution can be calculated using the formula: osmotic pressure = MRT, where M is the molar concentration of the solute, R is the ideal gas constant (0.0821 L·atm/(K·mol)), and T is the temperature in Kelvin. In this case, we have a 0.333 M solution of C6H12O6 (glucose). Plugging in the values, we get: osmotic pressure = (0.333 mol/L) [0.0821 (L·atm)/(K·mol)] (298 K) = 8.15 atm.

Answer 2

Answer is: the osmotic pressure of aqueous solution of aspirin is 0.0407 atm.

m(C₉H₈O₄) = 75 mg ÷ 1000 mg/g = 0.075 g.

n(C₉H₈O₄) = 0.075 g ÷ 180.16 g/mol.

n(C₉H₈O₄) = 0.000416 mol.

c(C₉H₈O₄) = 0.000416 mol ÷ 0.250 L.

c(C₉H₈O₄) = 0.00167 M; concentration of solution.

T(C₉H₈O₄) = 25°C = 298.15 K; temperature in Kelvins.

R = 0.08206 L•atm/mol•K; universal gas constant.

π = c(C₉H₈O₄) • T(C₉H₈O₄) • R.

π = 0.00167 mol/L • 298.15 K • 0.08206 L•atm/mol•K.

π = 0.0407 atm.