When 4.25 g of a nonelectrolyte solute is dissolved in 335 mL of water at 29 C,

Question

When 4.25 g of a nonelectrolyte solute is dissolved in 335 mL of water at 29 C, the resulting solution exerts an osmotic pressure of 859 torr. What is the molar concentration of the solution? How many moles of solute are in the solution? What is the molar mass of the solute?

Explanation / Answer

osmotic pressure = M * R * T where M is molarity, R is universal gas constant and T is temperature in Kelvin. R = 62.36 L torr K^-1 mol^-1, T = 273+29 = 302 K.------> Thus M = osmotic pressure /R * T = 859 / 62.36 * 302 = 0.0456 M. ------ 0.0456 mol is present in 1000 mL therefore in 335 ml moles of solute = 355*0.0456/1000 = 1.62x10^-2 mol -----> molar mass = mass of solute/ moles of solute = 4.25 g / 1.62x10^-2 mol = 262.3 g /mol.

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