Question

A solution of glucose in water is labelled as 10% w/w, what would be the molality and mole fraction of each component in the solution? If the density of solution is 1.2 g mL⁻¹, then what shall be the molarity of the solution?

Answer 1

Let mass of solution = 100 g

∴ Mass of glucose = 10 g

Mass of water = 100 – 10 = 90 g = 0.09 kg

No. of moles in 10 g glucose = `10/180`

= 0.0555 mol

No. of moles in 90 g H₂O = `90/18`

= 5 moles

Volume of solution = `(100  g)/(1.2  g  mL^(-1))`

= 83.33 mL

= 0.0833 L

Molality = `"Number of moles of solute"/"Mass of solvent in kg"`

= `(0.0555  mol)/(0.09  kg)`

= 0.617 m

x (Glucose) = `"Number of moles of solute"/"Number of moles of solution"`

= `0.0555/5.0555`

= 0.01

∴ x (H₂O) = 1 − 0.01 = 0.99

Molarity = `"Number of moles of solute"/"Volume of solution in L"`

= `0.0555/0.0833`

= 0.67 M

Answer 2

i. 10% w/w means that 100 g of the solution contains 10 g of glucose. Thus, the mass of water present is 100 − 10 = 90 g.

Moles of glucose present = `10/180`

= 0.055

Mass of solvent (water) = 90 g

= 0.090 kg

∴ Molality (m) = `0.055/0.090`

= 0.617 m

ii. Moles of water present = `90/18`

= 5

∴ Mole traction of glucose = `0.055/(0.055 xx 5)`

= 0.011

Mole fraction of water = `5/(0.055 xx 5)`

= 0.989

iii. Volume of 100 g solution = `"Mass"/"Density"`

= `100/1.2`

= 83.33 mL

Hence, molarity (M) of the solution

M = `(w xx 1000)/(M' xx v)`

= `(10 xx 1000)/(180 xx 83.33)`

= 0.67 M