The molality of an aqueous solution of a nonelectrolyte is 1.290 m?

Question

For pure water at 20 oC:
Po = 17.5 torr
kf = 1.86 oC/m
kb = 0.563 oC/m

1.)What is the boiling point of the solution to the nearest 0.1 oC?

2.)Assume that the molality and molarity of the solution are equal to estimate the osmotic pressure that is developed by the solution at 20.0 oC?

Answer 1

You just have to apply the respective formulas

The difference in boiling point is

dTb= kb*m =0.563*1.29= 0.726 oC but if you want 1 decimal accuracy it is 0.7 oC.

The boiling point of pure solvent (water) is 100.0 oC
Thus for the solution 100.0 + 0.7= 100.7 oC

Osmotic pressure π = CRT
but you said we must assume that C=m.
temperature is expressed in Kelvin so T=273+20= 293
we are expressing volume in lt so R =0.082 atm*lt/(mole*K)

Thus π = mRT = 1.29*0.082*293= 30.99 = 31 atm