Question

assuming the vapour pressure of water is increased from 30 to 31 mm hg at 25 c against an external pressure of 10 atm if the vapour pressure is to be maintained at 30mm hg for 1L of water at 25 c and 10 atm pressure how much NaCl in moles is to be added to the water?

Answer 1

Answer:

Nacl means Natrium Cloride and Natrium is a latin word

Answer 2

Concept:

The vapor pressure of any pure solvent decreases when a non-volatile solute is added to the solvent.

P* = xP'

where P* is the vapor pressure of solvent over the solution, P' is the vapor pressure of the pure solvent and x is the mole fraction of the solvent.

Given:

Vapor pressure of water at 25°C = 30 mm of Hg

The vapor pressure of the solvent is increased to 31 mm of Hg

Vapor pressure of pure solvent = 31 mm of Hg

Maintained Vapor pressure = 30 mm of Hg

Find:

The moles of NaCl are to be added to the water.

Solution:

P* = xP'

30 = x(31)

x = 0.96774

Mole fraction of solvent, water = 0.96774

Here, the water is the solvent and salt is the solute.

Mole fraction of solute + Mole fraction of solvent = 1

So, mole fraction of solute, salt (NaCl) = 1-x

= 1-0.96774 = 0.03226

The volume of water is 1 L.

So, Moles of Water, n = 1000/18 =  55.55 moles

Mole fraction of solvent = Mole of solvent / (Moles of solvent + Moles of solute)

(Moles of solvent + Moles of solute) = Mole of solvent/Mole fraction of solvent)

Moles of solute = (Mole of solvent/Mole fraction of solvent) - Moles of solvent

Moles of salt = (Moles of solvent)/(Mole fraction of solvent) - Moles of solvent

m = (55.55/0.96774) - (55.55) = 1.85 moles

Moles of NaCl to be added is the water is 1.85 moles.

Hence, 1.85 moles of NaCl must be added to the water.

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