A solution of Al_2(SO_4)_3 (d = 1.253 g/mL) | vedclass

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(A) Given: Density $(d)$ = $1.253 \ g/mL$,Weight percentage = $22\%$,Molar mass of $Al_2(SO_4)_3$ = $342 \ g/mol$.$1$. Molarity $(M)$ = $\frac{\% 	im

Vedclass · Accepted Answer

$0.805 \ M, 4.83 \ N, 0.825 \ m$

Vedclass · Answer

(A) Given: Density $(d)$ = $1.253 \ g/mL$,Weight percentage = $22\%$,Molar mass of $Al_2(SO_4)_3$ = $342 \ g/mol$.$1$. Molarity $(M)$ = $\frac{\% 	imes 10 	imes d}{	ext{Molar mass}} = \frac{22 	imes 10 	imes 1.253}{342} = 0.805 \ M$.$2$. Normality $(N)$ = Molarity $	imes$ n-factor. For $Al_2(SO_4)_3$,the n-factor is $6$ (total positive charge = $2 	imes 3 = 6$).$N = 0.805 	imes 6 = 4.83 \ N$.$3$. Molality $(m)$ = $\frac{	ext{mass of solute (g)} 	imes 1000}{	ext{Molar mass} 	imes 	ext{mass of solvent (g)}} = \frac{22 	imes 1000}{342 	imes (100 - 22)} = \frac{22000}{342 	imes 78} = 0.825 \ m$.

$A$ solution of $Al_2(SO_4)_3$ $(d = 1.253 \ g/mL)$ contains $22\%$ salt by weight. The molarity,normality,and molality of the solution are:

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