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A 0.15 M aqueous solution of KCl exerts ...

A 0.15 M aqueous solution of KCl exerts an osmotic pressure of 6.8 atm at 310 K. Calculate the degree of dissociation of KCl. (R = 0.0821 Lit. atm `K^(-1)"mol"^(-1)`).

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Concentration of solution `(C )=0.15 M`
Osmotic pressure `(pi) = 6.8` atm
Temperature (T) = 310 K
`R = 0.0821 "Lit atm K"^(-1)"mol"^(-1)`
Van.t Hoff factor (i) = ?
`pi =i.CRT`
`i=(pi)/(CRT)`
`=(6.8)/(0.15 xx 0.0821xx310)`
`i = 1.78`
Degree of dissociation,
`alpha=(i-1)/(m-1)`
m = Observed molar mass
`=2`
`alpha=(i-1)/(2-1) .alpha=i-1`
`=1.78-1=0.78`
Degree of dissociation `(alpha)=0.78 or 78%`
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