In Kjeldahi analysis, a flour sample weighing 0.9857 was digested in concentrated H,So, for 45 min. a concentrated solution of NaOH was added such that all the nitrogen was converted to NH. Following this, the NH, wes distilled into a fiask containing 50.00 ml of 0.1011 N H,So, The excess required 5.12 ml of 0.1266 N NaOH for titration. What is the % Nand Séprotein in the sample?

Chemistry
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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
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Show solution round off answer to four decimal places. Formula is given already Answer: % N = % protein =
N × V × mEq
sample weight(g)
% purity
sx wt (g)
х 100
N =
mEq x mL of solution
[(N x V)acid – (N × V) pase] × mEq
N = fM
% рurity :
-× 100
Sx wt (g)
(V blank – V sample) × N × mEq
sample weight (g)
% purity :
X 100
Transcribed Image Text:N × V × mEq sample weight(g) % purity sx wt (g) х 100 N = mEq x mL of solution [(N x V)acid – (N × V) pase] × mEq N = fM % рurity : -× 100 Sx wt (g) (V blank – V sample) × N × mEq sample weight (g) % purity : X 100
In Kjeldahl analysis, a flour sample weighing 0.9857 was digested in concentrated H;SO, for 45 min. a concentrated solution of
NaOH was added such that all the nitrogen was converted to NH3. Following this, the NH3 was distilled into a flask containing 50.00
mL of 0.1011 N H;SO. The excess required 5.12 ml of 0.1266 N NaOH for titration. What is the 9% Nand %protein in the sample?
Transcribed Image Text:In Kjeldahl analysis, a flour sample weighing 0.9857 was digested in concentrated H;SO, for 45 min. a concentrated solution of NaOH was added such that all the nitrogen was converted to NH3. Following this, the NH3 was distilled into a flask containing 50.00 mL of 0.1011 N H;SO. The excess required 5.12 ml of 0.1266 N NaOH for titration. What is the 9% Nand %protein in the sample?
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