Question

0.85% aqueous solution of NaNO₃ is apparently 90% dissociated at 27°C. Calculate its osmotic pressure. (R = 0.082 L atm K⁻¹ mol⁻¹)

Answer 1

NaNO₃ dissociates in solution as

	\[\ce{NaNO3 \phantom{..}<=> \phantom{.}Na+ + \phantom{.}NO^-_3}\]
Initially	1 mol                   -               -
At equilibrium	(1 − α) mol      (α mol)    (α mol)

∴ Total number of moles in solution = 1 − α + α + α

= 1 + α

van’t Hoff factor,

`i = "No. of moles in solution"/"No. of moles added"`

= `(1 + alpha)/1`

= 1 + α

Since the given salt dissociates 90% in the solution,

`alpha = 90/100`

= 0.9

Hence, for the given salt, i = 1 + a

= 1 + 0.9

= 1.9

For the given solution, mass of solute = 0.85 g

(normal molecular mass of NaNO₃ = 23 + 14 + 48 = 85)

∴ Moles of solute added = `0.85/85`

= 0.01

Volume of solution = 100 ml = 0.1 L

According to the modified equation of osmotic pressure,

π V =  i n RT

or `pi = (i n RT)/V`

= `(1.9 xx 0.01 xx 0.0821 xx 300)/0.1`

= 4.68  atm

Hence, the osmotic pressure of the given solution is 4.68 atm.