Question

When 2.13 g of a nonelectrolyte solute is dissolved in water to make 395 mL of solution at 21 °C, the solution exerts an osmotic pressure of 959 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 solution?

Answer 1

Osmotic Pressure = MRT

Where,

M = molarity in mol/L (molar concentration)

R = gas constant

T = temperature in Kelvin.

Given,

T = 21°C = 21+273K = 294K

Osmotic pressure = 959torr

1 torr = 0.00132atm

959 torr = 959*0.00132atm = 1.266atm

R = 0.0821latm/molK

Now,

Keeping the values in the formula,

1.266atm = M*0.0821latm/molK*294K

M = 1.266atm / 0.0821latm/molK*294K

M = 1.266atm / 24.14latm/mol

M (molar concentration)= 0.0524 mol/l

So, Molarity = moles of solute/ volume in L

Volume = 395ml = 395/1000L = 0.395L

moles of solute = Molarity*Volume in L

= 0.0524mol/L * 0.395L

moles = 0.021moles

Mass = 2.13g

Moles = Mass/Molar mass

Molar mass = Mass/Moles

= 2.13g / 0.021moles

Molar mass = 101.4g/mol