Chemistry!!!?
At the completion of distillation, the boric acid solution was titrated with hydrochloric acid (0.01N). A titration of 8.75mL of acid was required to reach the end point and a blank titration of 0.25mL was obtained.
Calculate the level of total volatile bases present in the fish flesh as mg N/100g flesh.
the total volatile bases mean nitrogen content and HVL 0.01N = 0.01mol/L
Answer:
the flesh take was 10 gms
the volume of extract was 10 ml and it was diluted to 50 ml
from this we are taking 10 ml for distillation,it is not necessary to include the volume of base and boric acid.
10gm --------10ml------50ml------ 10ml +base -----titration in boric acid medium.
the titre value of blank is 0.25ml
the nitrogen content is calculate as follows
(TVs - TVb) x 0.01N x 14 x1000
Nitrogen mg/L = __________________________
volume of sample taken
TVs = titre value for sample
TVs= titre value for blank,0.01N normality of HCl,
14 equivalent weight of nitrogen.1000 for milligram conversion.
(8.75-0.25)x 0.01 x14 x 1000
Nitrogen mg/L = ___________________________
10 ml
= 119mg
but we taken only 10 ml from the diluted part(50ml)
so we have to multiply this value by the factor of 5
5x 119mg = 595mg
this amount is actually present in 10 ml extract that is in 10gm fish sample.
we have express the result in N/100g
which mean amount of nitrogen present in 100g fish sample.
it is 595mg x 10 = 5950 mg or 5.95g present in 100g sample.
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