Chemistry, asked by Shirin35, 11 months ago

Lowering of vapour pressure due to a solute in 1 molal aqueous solution at 100degree celcious is

Answers

Answered by Anonymous
7

Answer:

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 = 13.44 torr

Hence, lowering of vapor pressure = 13.44 torr

Similar questions