`0.002 molar` solutiion of `NaCl` having degree of dissociation of `90%` at `27^(@)C` has osmotic pressure equal to a.0.94 bar , b.9.4 bar , c.0.094 bar , d.`9.4xx10^(-4)` bar

`0.002 molar` solutiion of `NaCl` having degree of dissociation of `90

1.	`0.002 molar` solutiion of `NaCl` having degree of dissociation of `90%` at `27^(@)C` has osmotic pressure equal to a.0.94 bar , b.9.4 bar , c.0.094 bar , d.`9.4xx10^(-4)` bar
Answer» c.`alpha=(i-1)/(m-1)` `0.9=(i-1)/(2-1)`,`i-1.9` Alternate method of calculate `(i)` `i=("Number of ions" xxalpha)+(1+alpha)` `=(2xx0.9)+(1-0.9)` [`alpha=90%` or `0.9`] `=1.8+0.1=1.9` `pi=iCRT` `=1.9xx0.002xx0.082xx300` `=0.094` bar

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`0.002 molar` solutiion of `NaCl` having degree of dissociation of `90%` at `27^(@)C` has osmotic pressure equal to a.0.94 bar , b.9.4 bar , c.0.094 bar , d.`9.4xx10^(-4)` bar

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