A2.4 m aqueous solution of an ionic compound with the formula mx2mx2 has a boiling point of 103.4 °c. calculate the van't hoff factor (i) for mx2mx2 at this concentration.

- Van't hoff factor(i) for the unknown ionic compound is 2.77.

Solution:- Elevation in boiling point is a colligative property and the equation used for solving this type of problems is...

delta Tb = i x m x kb

where delt Tb is elevation in boiling point.

i is Von't hoff factor.

m is molality of the solution and kb is the molal elevation constant. It's value for water is 0.512.

Boiling point of solution is given as 103.4 degree C and we know that water boils at 100 degree C.

So, delta Tb = 103.4 - 100 = 3.4 degree C

molality is given as 2.4m.

Let's plug in the values...

3.4 = i x 2.4 x 0.512

i = 3.4/(2.4 x 0.512)

i = 2.77

1 is the answer to this problem

c is having 4 valence electrons and is bonded to three hydrogens and one sulphur atom, therefore, the number of bond pair is equal to four. as all the valence electrons are used in forming the bonds so, there is no lone pair left on carbon.

s is having 6 valence electrons and is bonded to one other sulfur atom and one carbon atom, therefore, the number of bond pair is equal to 2. as only two valence electrons are used in forming the bonds so, four electrons are left. therefore, two lone pairs are present on s.