A student analysis nitrogen in blood by the kjeldahl method. He first standardiz

Question

A student analysis nitrogen in blood by the kjeldahl method. He first standardized his hydrochloric acid against sodium carbonate, then he standardized his NaOH against exact 25 mL aliquotes of the HCI. The organic nitrogen sample was digested, neutralized and distilled into 50, 000 mL of standardized HCI. and back titrated with standardized NaOH. calculate the average molarity of acid and base. calculate the average % N and standard deviation of the blood samples. calculate the confidence interval of the samples. Using the Grubb's test, determine whether any if the data can be rejected.

Explanation / Answer

Part-2:

1st blood sample(0.5007 g):

moles of NH3 formed = moles of NaOH reacted = MxV = 0.2045 mol/L x 0.02416 = 0.004941 mol

Hence moles of N atom in blood sample = moles of NH3 formed = 0.004941 mol

=> mass of nitrogen in blood sample = 0.004941 mol x 14 g/mol = 0.06917 g

% N = (0.06917 g / 0.5007 g) x 100 = 13.81 %

Similarly

2nd blood sample(0.4879 g):

moles of nitrogen = MxV =  0.2045 mol/L x 0.02363 L = 0.004832 mol

=> mass of nitrogen in blood sample = 0.004832 mol x 14 g/mol = 0.06765 g

% N = (0.06765 g / 0.4879 g) x 100 = 13.87 %

3rd blood sample(0.4217 g):

mass of nitrogen in blood sample = 0.2045 mol/L x 0.02735 L x 14 g/mol = 0.07830 g

% N = (0.07830 g / 0.4217 g) x 100 = 18.57 %

Hence average % N = (13.81% + 13.87% + 18.57 %) / 3 = 15.42 % (answer)

Standard deviation: 2.73 (answer)

Part-3: confidence interval = +/- 3.09

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