If the apparent degree of ionization of `KCl` in water at `290` K is `0.86`. Calculate the mass of `KCl` which must be made up to `1` `"dm"^(3)` of aqueous solution to the same osmotic pressure as the `4.0%` solution of glucose at that temperature.

Answer

8.9gm

Explanation

Glucose `("C"_(6)"H"_(12)"O"_(6)),`
Molar concentration of glucose solution. `("C"_(1))=(4.0xx1000)/(180xx100)=2/9"M"`
`pi_(1)="C"_(1)"RT"=2/9"RT"`

For `KCl` solution.
`"n"=2,alpha=0.86`
`"i"=1+("n"-1)alpha`
`"i"=1.86`
`w =` amount of `KCl` present in `1` dm`""^(3)` solution.
`pi_(2)=` Osmotic pressure of `KCl` solution
`pi_(2)=1.86xx("wRT")/(74.5xx1)`
`pi_(1)=pi_(2)`
`9/2"RT"=(1.86"wRT")/74.5`
`"w"=(2xx74.50)/(9xx1.86)`
`"w"=8.9"gm"`
`therefore` Mass of `KCl` present in `1` dm`""^(3)` of aq. solution `=8.9` gm