Question

Consider the six different solutions listed in the below.

Solution Number Type of Solution: 0.90 m BaCl2, 1.20 m BaCl2, 0.30 m KCl, 0.90 m KCl, 0.30 m glucose, and 1.20 m NaNO3

(a) Arrange these six solutions in order of decreasing osmotic pressures if the temperature of all solutions is 25°C.

(b) If the boiling-point elevations of these solutions were being analyzed in place of osmotic pressures, how would the order in part (a) change?

Answer 1

0.90m BaCl2 contains 2.70m particles since BaCl2 ionised into Ba⁺ and 2Cl^-

Similarly 1.20m BaCl₂ contains 3.60m particles

0.30m KCl will be ionised into K⁺ and Cl^- thus it contains 0.60 m particles

O.90m KCl will have 1.80m particles

Since as we know glucose does not lead to ionization hence it's 0.30 m contain 0.30m particles

1.20m NaNO₃ will be ionised into Na⁺ NO3^- hence it's solution will have 2.40m particles.

Where

is osmotic pressure

I = no. Of ions

M = molality

Since no. Of ion and molality increases hence osmotic pressure will also be increased.

So ,

i ) Order of decreasing osmotic pressure is

1.2m BaCl₂ >0.90m BaCl2 > 1.2m NaNO₃ > 0.90m KCl > 0.30mKCl > 0.30m glucose

ii)as we know boiling point elevation is colligative property and all colligative properties depend upon

Concentration of solute particles hence the order would be same for elevation in boiling point

1.2mBaCl₂ > 0.90m BaCl2 > 1.2m NaNO3 > 0.90mKC l>0.30mKCl > 0.30m glucose

Decreasing order of boiling point elevation