Question

Lowering of vapour pressure due to solute in 1 molal aqueous solution at 100°C is

BOOK ANSWER - 13.44 Torr

MY SOLUTION

Since molality is 1 mole kg^{​​​​​-1}

That's mean 1 mole of solute is present in 1 kg of solvent (H_{​​​​​​2}​​​​​O)

No. Of moles of Water = 1000g/18 = 55.55

No. Of moles of solute = 1

Mole Fraction of solute = 1/1+55.55 = 0.0176 .... (I)

Pressure of solvent (assuming ideal behaviour)

PV = nRT

P = 55.55 x 0.0821 x 373 / 1 (d of water = 1)

P = 1701.2774 atm

1 atm = 760 Torr

p_{​​​​​​solvent} = 1292970.8 Torr

We know from Raoult Law

∆p = X_{​​​​​​solute} . P(solvent)

∆p = 0.0176 x 1292970.8

∆p = 22756.286

So, According to my answer ∆p amount of vapour pressure will be low. But my answer and book answer is not same.

I don't want solution of this question.

You just tell me the place where I am going in wrong direction in my solution.

Answer 1

Dear student please check the given answer you will get the place where the wrong step you done.

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