Question

When 2.56 g of a nonelectrolyte solute is dissolved in water to make 825 mL of solution at 24 degrees Celcius, the solution exerts an osmotic pressure of 887 torr.

A) What is the molar concentration of the solution?

B) How many moles of solute are in the solution?

C) What is the molar mass of the solute in g/mol?

Answer #1

osmatic pressure = M * R *T

where, M = molarity

R = Gas constant

T = temperature = 24 + 273 = 297 K

osmatic pressure = 887 torr = 1.17 atm

1.17 = M * 0.0821*297

1.17 = M * 24.3837

M = 1.17 / 24.3837 = 0.04798

A) Therefore, molar concentration = 0.04798 M

M = number of moles / volume of solution in L

0.04798 = number of moles / 0.825

number of moles = 0.04798 * 0.825 = 0.03958

B) Therefore, the number of moles of solue = 0.03958 mole

C) number of moles = mass / molar mass

0.03958 mole = 2.56g / molar mass

molar mass = 2.56g/ 0.03958 mole = 64.68 g/mol

Therefore, molar mass = 64.68 g/mol

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