`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

Explanation

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