Question

If 5g NaOH dissolved in 50g of water to reduce the pressure having vapour pressure of pure water is 630mmHg find relative lowering of vapour pressure

Answer 1

weight of NaOH = 5g

molecular weight of NaOH = 40g/mol

so, number of mole of NaOH , n = 5/40 = 0.125

weight of water = 50g

molecular weight of water = 18g/mol

so, number of mole of water, N = 50/18 = 2.77

now using formula,

relative lowering of vapor pressure = mole fraction of solute

= n/(n + N)

= 0.125/(0.125 + 2.77)

= 0.125/2.895

= 0.0431

and lowering of vapor pressure = vapor pressure of pure water × relative lowering of vapor pressure

= 630 mm Hg × 0.0431

= 27.153 mm Hg

Answer 2

This Question can be easily done by using Roult's law.

According to the Roult's law, if an non-volatile solute is dissolved in an liquid solvent which should be volatile, then Relative lowering of vapor pressure of the solvent is directly proportional to the mole fraction.

Relative lowering of vapor pressure = Mole Fraction of the solute.

Mole Fraction of the solute = Moles of solute/Total moles in solution.

 = (5/40)/(5/40 + 50/18)

 = (5/40)/2.902

 = 0.043

Now, If you want to calculate the exact Vapor pressure of the solvent after the solute has been added, then,

(P - Ps)/P = Mole Fraction of solute,

where P is the Vapour pressure of Pure solvent, and Ps is the Vapour pressure of solution.

  630 - Ps = 0.043 × 630

  Ps = 630 - 27.129

  Ps = 602.87 mm of Hg.

Hope it helps.