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

Answer» Normal osmotic pressure`=W/(MwxxV)xxRxxT`
When no dissociation has taken place
`W/(Mw)=0.1`,`V=1 L`,`R=0.821`,`T=300 K`
Normal osmotic pressure =`0.1/1xx0.0821xx300`
`=2.463 atm`
Potassium ferrocyanide is an electrolyte. It dissociates as
`K_(4)[Fe(CN)_(6) hArr 4 K^(o+) + [Fe(CN)_(6)]^(4-)`

Total numberof particles =`1-alpha+4alpha=1+4alpha`
`alpha=0.46`,so `1+4alpha=1+4xx0.46=2.84`
`"Observed osomotic pressure"/"Normal osomotic pressure"=2.84/1`
Observed osmotic pressure =`2.84xx2.463=6.995`


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