Question

100. mL of solution and is found to have an osmotic pressure of 28.1 mmHg at 20°C. Calculate the molar mass of the compound. (1 atm 760 mmHg) 1 0.102 g of an unknown non electrolyte solute is dissolved in enough water to make A. 663 g/mol B. 0.872 g/mol C. 1.15 g/mol D. 727 g/mol E. 1.10 x 102 g/mol

Answer 1

Answer : A) 663 g/mol

osmatic pressure = 28.1 mmHg = 0.03697 atm

temperature = 20 oc = 293 K

volume = 100 mL = 0.1 L

osmatic pressure = i x M x S x T

0.03697 = 1 x M x 0.0821 x 293

M = 1.537 x 10^-3 M

Molarity = moles / volume

1.537 x 10^-3 = moles / 0.1

moles = 1.537 x 10^-4

moles = mass / molar mass

1.537 x 10^-4 = 0.102 / molar mass

molar mass = 663 g/mol