What will be the osmotic pressure of a 0.03 | vedclass

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(A) The dissociation of Aluminium sulphate is: $Al_2(SO_4)_3 \rightarrow 2Al^{3+} + 3SO_4^{2-}$.Here,the number of ions produced per formula unit is $

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$0.56$

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(A) The dissociation of Aluminium sulphate is: $Al_2(SO_4)_3 ightarrow 2Al^{3+} + 3SO_4^{2-}$.Here,the number of ions produced per formula unit is $n = 5$.The van't Hoff factor $i$ is calculated as: $i = 1 + (n-1)\alpha = 1 + (5-1) 	imes 0.9 = 1 + 3.6 = 4.6$.Given $Normality = 0.03 \ N$ and the $n$-factor for $Al_2(SO_4)_3$ is $6$ (total positive charge),the Molarity $M$ is: $M = \frac{Normality}{n	ext{-factor}} = \frac{0.03}{6} = 0.005 \ M$.The osmotic pressure $\pi$ is given by: $\pi = iCRT$.Substituting the values: $\pi = 4.6 	imes 0.005 	imes 0.0821 	imes 300 \ K$.$\pi = 0.566 \ atm$.

What will be the osmotic pressure of a $0.03 \ N$ solution of Aluminium sulphate at $27 \ ^\circ C$? If the salt dissociation in the solution is $90 \ \%$,the osmotic pressure is . . . . . . $atm$.

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