For the reaction: (NH2)2CO --------> (NH4)2 CO3 ----------> 2NH3 + CO2 +H20 The
ID: 521422 • Letter: F
Question
For the reaction: (NH2)2CO --------> (NH4)2 CO3 ----------> 2NH3 + CO2 +H20
The standard required 20 mL of 0.01N NaOH to titrate the HClThe 1 mL sample required 10 mL of 0.01N NaOH to titrate the HClWhat was the concentration of NH3 in the sample (mg/mL)?, and what was the original concentration of urea if the urine sample was diluted 10 times? Given: Mw Urea: 60.0 g/mol; Mw NH3 =17.01 g/mol ; Stoichiometry: (1 Urea/2 NH3)!
a) 1.7 mg/mL, 60 mg/mL
b) 1.7 mg/mL, 30 mg/mL
c) 17 mg/mL, 60 mg/mL
d) 17 mg/mL, 30 mg/mL
Explanation / Answer
for the given reaction,
moles of NaOH used = moles of HCl = moles of NH3
moles of NaOH = 0.01 M x 10 ml = 0.1 mmol for 1 ml sample
concentration of NH3 in the sample = 0.1 mmol x 17.01 g/mol/1 ml
= 1.7 mg/ml
when the urine sample is diluted 10 times,
concentration of NH3 = 1.7 x 10 = 17 mg
moles NH3 = 17/17 = 1 mmol
original concentration of urea = 1 mmol x 60 = 60 mg/ml
Answer : a) 1.7 mg/ml, 60 mg/ml
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