2) For the reaction: (NH2)2CO > (NH4)2 CO3 > 2NH3 + CO2 +H20 The standard requir

Question

2) For the reaction: (NH2)2CO > (NH4)2 CO3 > 2NH3 + CO2 +H20 The standard required 20 mL of 0.01N NaOH to titrate the HCl The 1 mL sample required 10 mL of 0.01N NaOH to titrate the HCl What 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

Moles of HCl present before reaction = 20 x 0.01 /1000 = 0.0002 Moles

Moles of HCl present after neutralizing ammonia = 10 x 0.01 /1000 = 0.0001 Moles

Moles of HCl used to  neutralizing ammonia = 0.0002 -0.0001 = 0.0001 Moles

Concentration of ammonia solution = 0.0001 x 17 = 0.0017 g/ml or 1.7mg /ml

Concentration of urea = 0.0005 x 60 = 30 mg/ml

Answer is option b 1.7 mg/mL, 30 mg/mL

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