Question

A solution is prepared from 4.5750 g of magnesium chloride and 43.248 g of water. The vapor pressure of water above this solution is found to be 0.3626 atm at 348.0 K. The vapor pressure of pure water at this temperature is 0.3804 atm. Part A Find the value of the van't Hoff factor i for magnesium chloride in this solution. Express your answer using four significant figures.

Answer 1

We kone that,

Molar mass of MgCl​​​2 =95.211g/mol

Molar mass of H2O = 18.0153g/mol

Given data

Mass of MgCl​​​2 =4.5750g

Mass of H2O = 43.248g

Step(1) Calculation for number of mole

Number of mol = Mass/molar mass

Number of mole of

MgCl2 (n​​​​​​1 ) = 4.5750g/(95.211g/mol)

n​​​​​​1 = 0.048mol

H2O (n​​​​​​2​) = (43.248g)/(18.0153g/mol)

n​​​​​​2 = 2.401mol

.

.

P° = vapour pressure of pure Water =0.3804atm

P = vapour pressure of solution = 0.3626atm

Relative lowering in vapour pressure will be given as follows

∆P/P° = i(mole fraction of MgCl​​​2) ......(1)

∆P = P° - P = (0.3804 atm) - (0.3626atm)

∆P = 0.0178atm

i = wan't Hoff factor

mole fraction of MgCl​​​2 = (n​​​​​​1)/(n​​​​​1+n​​​2) .........(2)

Step (2) Calculation for mole fraction of MgCl​​​2

Putting the values of mole in equation Number (2)

Mole fraction of MgCl​​​2 = (0.048)/(0.048+2.401)

Mole fraction of MgCl​​​2 = (0.048)/(2.449)

Mole fraction of MgCl​​​2 = 0.0196

Step(3) Calculation for van't hoff factor

Putting the all value in equation Number (1)

(0.0178atm)/(0.3804atm) = i(0.0196)

0.0468 = i(0.0196)

i = 0.0468/0.0196

i = 2.387

Hence , van't hoff factor = 2.387