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