Question

A decinormal solution of sodium chloride exerts an osmotic pressure of 4.82 atmosphere at 27°C. Calculate the degree of dissociation of sodium chloride.

Answer 1

Given: π = 4.82 atmosphere

T = 27°C = 27 + 273 = 300 K

C = 0.1

R = 0.821 L atm K⁻¹ mole⁻¹

π = i CRT

or `i = pi/(CRT)`

= `4.82/(0.1 xx 0.821 xx 300)`

= 0.195

	NaCl →	Na+	Cl−
Initially	1	0	0
At equ.	1 − α	α	α

Total moles at equilibrium = 1 − α + α + α

= 1 + α

`i = "Observed no. of moles"/"Normal no. of moles"`

`i = (1 + alpha)/1`

i = 1 + α

0.195 = 1 + α

α = 1 − 0.195

α = 80.5

α = 80.5%