Question

An aqueous solution is made by dissolving 10 g of glucose (C₆H₁₂O₆) in 90 g of water at 300 K. If the vapour pressure of pure water at 300 K is 32.8 mm Hg, what would be the vapour pressure of the solution?

(Given: C = 12, H = 1 and O = 16)

Answer 1

Given: Mass of glucose (C₆H₁₂O₆) = 10 g

Mass of water (H₂O) = 90 g

Vapour pressure of pure water at 300 K = 32.8 mm Hg

Atomic masses:

C = 12, H = 1, O = 16

First, we will calculate the molar mass of glucose and water,

Molar mass of glucose (C₆H₁₂O₆):

= (6 × 12) + (12 × 1) + (6 × 16)

= 72 + 12 + 96

= 180 g/mol

Molar mass of water (H₂O):

= (2 × 1) + (1 × 16)

= 2 + 16

= 18 g/mol

Now, we will calculate the moles of glucose and water,

Moles of glucose = `"Mass of glucose"/"Molar mass of glucose"`

= `10/180`

= 0.0556 moles

Moles of water = `"Mass of water"/"Molar mass of water"`

= `90/18`

= 5.00 moles

Now, we will calculate the mole fraction of water,

Χ_water = `"Moles of water"/("Moles of water" + "Moles of glucose")`

= `5.00/(5.00 + 0.0556)`

= `5.00/5.0556`

= 0.989

Now, using Raoult’s law, we will calculate the vapour pressure of the solution:

`P_"solution" = chi_"water" xx P_"water"^0`

= 0.989 × 32.8

= 32.4 mm Hg

Thus, the vapour pressure of the glucose solution at 300 K is 32.44 mm Hg.