Answer:
82.53 % protein in the milk.
Explanation:
Titration formula:
Conc (acid) × Vol (acid) = Conc (base) × Vol (base)
Beginning from back-titration:
0.5 M HCl × Vol (HCl) = 0.3512 (conc of NaOH) × 34.50 mL (vol of NaOH)
= [tex]\frac{12.1164}{0.5}[/tex] = 24.2328 mL
⇒ Vol of HCl used in initial titration is 24.2328 mL
So from Initial Titration:
Using same titration formula,
0.5 × 24.2328 = Conc. (base) × 100 ml
Conc (base) = 12.1164 ÷ 100
= 0.121164 M.
But concentration = mass ÷ molar mass
0.121164 = [tex]\frac{100 g}{ molar mass}[/tex]
= 825.3277 g/mol of protein
⇒ [tex]\frac{825.32765}{1000}[/tex] × 100
= 82.53 % protein in the milk.
