An M/10 solution of potassium ferrocyanide is `46%` dissociated at `300 K`. What will be its osmotic pressure?

Explanation

Normal osmotic pressure `= (w)/(m xx V)xx S xx T`
(when no dissociation has taken place)
`(w)/(m)=0.1,V = 1 litre, S=0.0821, T=18+273 = 291 K`
Normal osmotic pressure `= (0.1)/(1)xx0.0821 xx 291=2.389` atm
Potassium ferrocyanide is an electrolyte. It dissociates as :
`underset((1-alpha))(K_(4)Fe(CN)_(6))hArrunderset(4alpha)(4K^(+))+underset(alpha)([Fe(CN)_(6)]^(4-))`
Total number of particles `= 1 - alpha + 4 alpha + alpha = 1+4 alpha`
`alpha =0.46` , so, `1 + 4 alpha = 1+4 xx 0.46 = 2.84`
`("Observed osmotic pressure")/("Normal osmotic pressure")=(2.84)/(1)`
Observed osmotic pressure `=2.84xx2.389`
`=6.785` atm.