Calculate the vapour pressure of solution formed by dissolving 5g of Glucose in 500g of Water. Vapour pressure of Pure Water = 760 mm
Answers
Explanation:
Answer:Mole fraction of solute, x 2 = p 1∗ −Δp = 30003000−2985 =0.005Now, x 2 = M 1 m 1 + M 2 m 2 m 2 /M 2 ≈ m 1 /M 1 m 2 /M 2 ∴0.005= 100/185/M ... More
Given:
The mass of glucose = 5 grams
The amount of water = 500 grams
The vapor pressure of pure water = 760 mm of Hg
To Find:
The vapor pressure of the solution.
Solution:
The number of moles of solute (glucose) = Given mass/ Molecular mass
⇒ The number of moles of glucose (n₁) = 5g/180gmol⁻¹ = 0.0277 moles
The number of moles of solvent (Water) (n₂) = 500g/18gmol⁻¹ =27.77mole
The relative lowering in the vapor pressure,
⇒ (p° - p)/ p° = n₁/n₂
Where p° = Vapor pressure of the solvent
p = Vapor pressure of the solution
On rearranging the equation,
p = p° ( n₂ - n₁) / n₂
⇒ p = 760 ( 27.77 - 0.0277) / 27.77
⇒ p = 759.24 mm of Hg
Hence, the vapor pressure of the solution is 759.24 mm of Hg.