`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

Options

A. 0.94 bar
B. 9.4 bar
C. 0.094 bar
D. `9.4xx10^(-4)` bar

Answer

C

Explanation

`alpha = (i-1)/(n-1)`
`0.9=(i-1)/(2-1), i=1.9`
Alternate method to calculate (i)
`i = ("Number of ions" xx alpha)+(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