Question

Calculate the molarity and normality of solution that contains 16.2g of salt fe(so4) in 200ml of solution

Answer 1

Answer:

molar mass of Fe2(SO4)3 is 56 x 2 + 3 x 32 + 12 x 16

=> 112 + 96 + 192 => 400 grams per mole.

So, 16.2 grams of Fe2(SO4)3 is 16.2/400 => 0.0405 moles of Fe2(SO4)3 .

Volume of solution is 200 mL or 0.2 liters.

Molarity =  (moles of solute) / (liters of solution)

=> 0.0405/0.2

=>0.2025 M (molar)

Hence, molarity of solution is 0.2025 M (molar).

Now, we have to find normality,

Looking at the formula of Fe2(SO4)3 , we can tell that 2 Fe3+ ions will be formed from each molecule of Fe2(SO4)3.

N-factor for a salt is the total number of H+ ions that can completely replace the positive cation present in it.

Here, 2Fe3+ ions are present, so the N-Factor is 2 x 3 => 6 for Fe2(SO4)3

Hence, normality = molarity x n-Factor

normality = 0.2025 x 6 => 1.2150 N

Hence, normality for this solution is 1.215 N.

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