Question

calculate the amount of f.a.s required to prepare M/20 standard solution of f.a.s​

Answer 1

Answer:

The amount of f.a.s. required to prepare M/20 standard solution of f.a.s. is 19.6 g.

Concept:

Molarity:  Molarity is defined as the number of moles of solute dissolved per unit volume of the solution. Mathematically, it can be expressed as,

M = n/V

where M = Molarity of the solution

n = Number of moles of solute

V = Volume of the solution

The number of moles of solute is given by,

n = m/W

where m = Mass of the solute in grams

W = Molar mass of the solute

Given:

Molarity of f.a.s. (ferrous ammonium sulphate) solution, M = M/20 = 1/20 M

Find:

The amount of f.a.s. (ferrous ammonium sulphate) required to prepare the solution.

Solution:

Here solute is ferrous ammonium sulphate (or Mohr's salt) [(NH₄)₂ Fe (SO₄)₂ · 6H₂O].

Molar mass of solute, W = 2(Molar mass of NH₄) + Molar mass of Fe + 2(Molar mass of SO₄) + 6(Molar mass of H₂O)

Molar mass of solute, W = 2(18) + 56 + 2(96) + 6(18)

                                        = 36 + 56 + 192 + 108

                                        = 392 g/mol

Let the mass of solute in grams be w.

Now, molarity is given 1/20 M. This means that 1/20 moles are dissolved per unit volume.

Number of moles of solute, n = 1/20

But n = m/W

∴ m/W = 1/20

m = W/20

m = 392/20

m = 19.6 g

Amount of f.a.s., m = 19.6 g

Hence, the amount of f.a.s. required to prepare M/20 standard solution of f.a.s. is 19.6 g.

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