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weeeeeb [17]
2 years ago
6

8. The protein in a 1.2846-g sample of an oat cereal is determined by a Kjeldahl analysis. The sample is digested with H2SO4, th

e resulting solution made basic with NaOH, and the NH3 distilled into 50.00 mL of 0.09552 M HCl. The excess HCl is back titrated using 37.84 mL of 0.05992 M NaOH. Given that the proteins in grains average 17.54% w/w N, report the %w/w protein in the sample.
Chemistry
1 answer:
docker41 [41]2 years ago
8 0

Answer: The %w/w protein in the sample is 15.2 %

Explanation:

To calculate the number of moles for given molarity, we use the equation:

\text{Moles of solute}={\text{Molarity of the solution}}\times{\text{Volume of solution (in L)}}    

\text{Moles of} HCl={0.09552\times 50.00}{1000}=0.0047moles

\text{Moles of} NaOH={0.05992\times 37.84}{1000}=0.0023moles

HCl+NaOH\rightarrow NaCl+H_2O  

According to stoichiometry :

1 mole of NaOH require 1 mole of HCl

Thus 0.0023 moles of NaOH will require=\frac{1}{1}\times 0.0023=0.0023moles  of HCl

moles of HCl used = (0.0047-0.0023) = 0.0024

NH_3+HCl\rightarrow NH_4Cl

1 mole of HCl uses = 1 mole of ammonia

Thus 0.0024 moles uses = \frac{1}{1}\times 0.0024=0.0024moles of ammonia

Mass of ammonia= moles\times {\text {Molar mass}}=0.0024\times 17g/mol=0.0408g

17 g of ammonia contains = 14 g of Nitrogen

Thus 0.0408 g of ammonia contains = \frac{14}{17}\times 0.0408=0.034 g of Nitrogen

Now 17.45 g of Nitrogen is present in = 100 g of protein

Thus 0.034 g of Nitrogen is present in =\frac{100}{17.45}\times 0.034=0.195g of protein

Now % w/w of protein = \frac{0.195}{1.2846}\times 100=15.2\%

Thus %w/w protein in the sample is 15.2%

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Since the cost of materials cannot be affected and have to be bought under similar rate of cost of working or process can be decreased. By using  machines calculation and mechanical devices this come would be accurate and the laws would be minimised. Hence the companies outcome would increase.

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