The Grass Doctor

Urea, (NH2)2CO is a commonly used fertilizer. if a 3.42 m (aq) of urea has a density of 1.045 g/mL...?

calculate urea in terms of a) Molarity b)weight percent c) mole fraction

Public Comments

1. 3.42 m means 3.42 moles in 1 Kg of water moles water = 1000 g /18.02 g/mol=55.5 mole fraction urea = 3.42 / 55.5 + 3.42=0.0580 mass urea = 3.42 mol x 60.055 g/mol=205.4 g % by mass = 205.4 x 100 / 1000 + 205.4=17.0 mass solution = 1205.4 g Volume solution = 1205.4 / 1.045=1153 mL M = 3.42 mol/ 1.153 L=2.96
2. Atomic weights: C=12, H=1 O=16 N=14 CH4N2O=60 Let urea be called U. Let urea solution be called S. 3.42molU/1kgH2O x 60gU/1molU = 205.2gU/1kgH2O So there is 205.2g urea in 205.2 + 1000 = 1205.2g solution 205.2gU/1205.2gS x100% = 17.0 wt-% urea [answer to b)] 1.045gS/1.000mLS x1000mLS/1LS x 17.0gU/100gS x 1molU/60gU = 2.96 mole/L [molarity, the answer to a)] 1000gH2O x 1molH2O/18gH2O = 55.56 moles H2O 55.56 + 3.42 = 58.98 total moles 3.42/58.98 = 0.05799 mole fraction urea, so 0.942 mole fraction H2O [answer to c)]