Determine the molarity and nomality of the aqueous solution containing 100g of Al2(SO4)2?

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Determine the molarity and nomality of the aqueous solution containing 100g of Al2(SO4)2?

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Answer 1 · 2017-08-06T13:27:11

$Molarity = \frac{Moles of solute}{Volume of solution}$ $"Molarity"="Moles of solute"/"Volume of solution"$

Explanation:

Now $moles of solute = \frac{100 \cdot g}{342.15 \cdot g \cdot m o l^{- 1}} = ? ? \cdot m o l$ $"moles of solute"=(100*g)/(342.15*g*mol^-1)=??*mol$

But we ain't got a volume of solvent specified........so we cannot proceed......

And by the way $aluminum sulfate$ $"aluminum sulfate"$ is $A l_{2} {(S O_{4})}_{3}$ $Al_2(SO_4)_3$ , the salt of $A l^{3 +}$ $Al^(3+)$ and $S O_{4}^{2 -}$ $SO_4^(2-)$ ions..........

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Determine the molarity and nomality of the aqueous solution containing 100g of Al2(SO4)2?

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