Question

When 3.34 g of a nonelectrolyte solute is dissolved in water to make 515 mL of solution at 24 °C, the solution exerts an osmotic pressure of 955 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?

Answer 1

Osmotic pressure π=iMRT

Where π= osmotic pressure=955 torr, i=Vant Hoff's factors=1 for nonelectrolyte.

M= molar concentration=moles/volume

R=gas constant=62.363 L.torr.mol^-1.K^-1.

T=temperature=24°C=24+273 K=297 K.

Substitute all these values in above equation

π=iMRT

955 torr=1 x M x 62.363 L.torr.mol^-1.K^-1 x 297 K

M=955/(18521.8 L.mol^-1)=0.0515 mol/L

Therefore molar concentration=0.0515 M.

Given volume of solution=515 mL=0.515 L (1L=1000 mL)

We know that Molar concentration=moles/volume

Moles=Molar concentration x volume=0.0515 mol/L x 0.515 L

Number of moles of solute in solution=0.0265 mol.

We know that moles=mass/molar mass

Given mass of nonelectrolyte=3.34 g

Molar mass=mass/moles=3.34 g/0.00265 mol

The molar mass of solute=125.93 g/mol.