Question

In the lab, you prepare 50 mL of FeCl₃*6H₂O 0.1 M. Now find the Normality [N], %(w/v) and ppm.

Answer 1

Sol:-

• 1 L = 1000 mL
• Atomic mass of Fe = 56 g/mol.
• Atomic mass of Cl = 35.5 g/mol.
• Atomic mass of H = 1 g/mol.
• Atomic mass of O = 16 g/mol.
• Molecular mass of FeCl₃•6H₂O = {56 + (3*35.5) +6(2*1 + 16 )} = { 56 + 106.5 + 6(18)} g/mol. = {56 + 106.5 + 108} g/mol. = 270.5 g/mol
• Molecular mass of FeCl₃(soluts) = { 56 + (3*35.5)} g/mol. = {56 + 106.5} g/mol. = 162.5 g/mol
• n-factor in FeCl₃ = 3
• n-factor = number of e^- transfer in reaction
• Density of FeCl₃•6H₂O = 1.82 g/mL

Data provided in the question is :-

• Volume of solution = 50 mL = 0.050 L
• Concentration of solution(Molarity) = 0.1M

Formula used :-

• Normality = (gram-equivalent of Solute) /(Volume of solution in litres)
• Gram-equivalent = (weight)/(equivalent weight)
• Equivalent weight = (molecular weight)/(n-factor)
• Normality = (n-factor) *(molarity)
• Mole of solute = (molarity) * (volume of solution in litres)
• Mass of solute =( mole of solute) *(molecular mass of solute)
• %(w/v) = {(mass of solute)/(volume of solution)} * 100
• ppm = {(mass of solute) /(mass of solution) } * 10⁶

Calculation :-

Moles of solute = (molarity) * (Volume of solution in litres)

= (0.1M) * (0.050 L) mol.

= 0.0050 mole

Mass of solute(FeCl₃) = (mole of FeCl₃) * ( molecular mass of FeCl₃)

= (0.0050 mole) * (162.5 g/mol.)

= 0.8125 g

Equivalent weight of solute(FeCl₃) = (molecular weight of FeCl₃)/ (n-factor)

= (162.5)/ 3

= 54.166 g.

Gram-equivalent of solute(FeCl₃) = (weight of FeCl₃)/(equivalent weight of FeCl₃)

=( 0.8125 g) /(54.166 g)

= 0.0150

Normality = (gram-equivalent of FeCl₃)/(Volume of solution in litres)

  = (0.015 )/(0.050 L)

= 0.3 gram-equivalent/L

%(w/v) = {(weight of solute(FeCl₃))/(Volume of solution) }*100

= {(0.8125 g) /(50 mL) }*100

= (0.01625 g/mL) * 100

= 1.625 %

Mass of solution(FeCl₃•6H₂O) = (density of FeCl₃•6H₂O) * (Volume of solution(FeCl₃•6H₂O))

= (1.82 g/mL) * (50 mL)

= 91 g

ppm =

{(mass of solute(FeCl₃))/(mass of solution(FeCl₃•6H₂O))} * 10⁶

= [{(0.8125 g) /(91 g) } * 10⁶] ppm

= [0.008928 * 10⁶ ]ppm

= 8.928 * 10³ ppm

Normality = 0.3 gram-equivalent/L

%(w/v) = 1.625 %

ppm = 8.928 * 10³ ppm