Question

Estimate the lowering of vapour pressure due to the solute (glucose) in a 1M aqueous solution at 100°C

Answer 1

Answer:

13.45 torr

Explanation:

First of all find mole fraction of gas . 

Let mole fraction of gas is x 

we know, the relation between mole fraction and molality

molality = x × 1000/(1 - x)M 

Here, M is Molecular weight of solvent . 

for aqueous solution , solvent is water . 

∴ Molecular weight of water , M = 18g/mol 

Now, 1 = x × 1000/(1 - x) × 18

⇒18(1 - x) = 1000x 

⇒18 = (1000 + 18)x 

⇒x = 18/1018 

Now, use formula , 

Relative lowering of vapor pressure = mole fraction of gas 

∆P/P₀ = x 

Here P₀ is initial pressure ,at STP , P₀ = 760 torr 

so, ∆P = 760 × 18/1018 = 13.45 torr 

Hence, lowering of vapor pressure = 13.45 torr