Question

The mole fraction of glucose in an aqueous solution is 0.015. The density of the solution is 1.05 g/mL. Calculate the molarity and molality of the solution.

Answer 1

The molality and molarity of a solution is given as:

Molality = Moles of solute / Mass of solvent (in Kg) ....(A)

Molarity = Moles of solute / Volume of solution (in Litre) ....(B)

We are given with,

Mole fraction of glucose = 0.015

Density of solution = 1.05 g/mL

The mole fraction of glucose is given as:

Mole fraction of glucose = Moles of glucose / Moles of glucose + Moles of water

Since mole fraction is always equal to 1, so the above equation becomes,

=> 0.015 = Moles of glucose / 1

=> Moles of glucose = 0.015

=> Moles of water = 1 - 0.015 = 0.985

Now mass of water and mass of glucose can be calculated as:

Mass of water = Moles of water * Molar mass of water

=> Mass of water = 0.985 * 18 = 17.73 g = 0.01773 Kg

Mass of glucose = Moles of glucose * Molar mass of glucose

=> Mass of glucose = 0.015 * 180 = 2.7 g

Now molality can be calculated from the equation (A),

Molality = Moles of glucose / Mass of solvent (in Kg)

=> Molality = 0.015 / 0.01773 = 0.846 m

We know,

Density = Mass / Volume

=> Density of solution = Mass of glucose + Mass of water / Volume of solution

=> 1.05 g/mL = 2.7 g + 17.73 g / Volume of solution

=> Volume of solution = 20.43 / 1.05 = 19.457 mL = 0.0195 L

Similarly molarity can be calculated from the equation (B),

Molarity = Moles of glucose / Volume of solutioon (in L)

=> Molarity = 0.015 / 0.0195 = 0.769 M