Question

Calculate the molarity and normality of a solution that contains 16.2g of the salt Fe2(SO4)3 in 200ml of solutin.

Answer 1

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.

Answer 2

Answer:

M = 0.024M(molar) and N = 1.49N

Explanation:

given:

m=16.2g

V=200ml(0.2L)

charge = Fe+3, SO4-2 => 3e*2e = 6e

M=2*18.998 + (32.06+4*(15.999))*3 = 326.164 g/mol -> 2*Fe + (S+4*O)*3

required:

A) M=?

B) N=?

#Lets Find The Mole

mole=given mass/Molar mass

mole = 16.2 g / 326.164 g/mol

mole=0.049 mol

#Solution

A) Molarity(M) =  Mole of solute/Volume(Liter)

M = 0.049mol / 0.2 L

M = 0.024M(molar)

B) Normality(N) = (given mass * charge) / Molar mass * Volume(Liter)

N = (16.2 g*6e) / 326.164 g/mol * 0.2L

N = 97.2/65.23

N = 1.49N