0.85% aqueous solution of NaNO_3 is apparently 90% dissociated. The osmotic pressure of solution at 300 K is

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Correct answer: (A) 4.674 atm. M=(W×1000)/(mol.wt.×V) Molarity =(1000cm^3×0.85g)/(100cm^3×85gmol^(-1))=0.1molL^(-1) NaNO_3→Na^(+)+NO3- So i=(1-α+α-α)/1 =1+α=1+0.90=1.90 π=iCRT=1.90×0.1×0.082×300 =4.674 atm.

0.85% aqueous solution of $N a N O_{3}$ is apparently 90% dissociated. The osmotic pressure of solution at 300 K is

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0.85% aqueous solution of NaNO3 is apparently 90% dissociated. The osmotic pressure of solution at 300 K is

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0.85% aqueous solution of $N a N O_{3}$ is apparently 90% dissociated. The osmotic pressure of solution at 300 K is