Question

A solution is prepared by dissolving 5g of non volatile solute in 95 g of water. It has a vapour pressure of 23.375 mm hg at 25°c.Calculate the molar mass of the solute. (vp of pure water at 25°c is 23.75mm hg)

Answer 1

Answer:

Explanation:

Rlvp= mole fraction of solute

Answer 2

Answer:

Explanation:

Quantity of water = 95g (Given)

Vapour pressure = 23.375mm (Given)

Temperature =  25°c (Given)

According to the Raoult’s Law -

p0 - p/p0 = w × M/ m× W

where -

p0 is the vapour pressure of pure water  

p is the vapour pressure of solvent  

w is the weight of solute  

m is the molecular weight of solute  

W is the molecular weight of solvent  

M is the Molar Mass of Water = 18 g mol -1  

Substituting the values in the formula, we will get -

= 23.75 - 23.375 / 23.75 = 5 × 18/ m × 95

= 23.75 - 23.375 / 23.75 = 90/95m

= 60m/s