1. Wine is approximately 12% ethanol C2H5OH by volume . Ethanol has a molar mass
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Question
1. Wine is approximately 12% ethanol C2H5OH by volume . Ethanol has a molar mass of 46.06g/mol and a density 0.789g/mL. How many moles of ethanol are present in a 750mL bottle of wine ?
2. A 50.0g sample of industrial wastewater was determined to contain 0.48 mg of mercury. Express the mercury concentration of the waste water in ppm and ppb units.
3. The molecular formula for a compound is CX4. If 2.819 g of this compound contains 0.102 g of carbon, what is the atomic weight of X?
4. A 4.628-g sample of an oxide of iron was found to contain 3.348 g of iron and 1.280 g of oxygen. What is simplest formula for this compound?
5. Aluminum oxide, Al2O3, is used as a filler for paints and varnishes as well as in the manufacture of electrical insulators. Calculate the number of moles in 47.51 g of Al2O3.
Explanation / Answer
1) Total volume of wine = 750ml
volume of ethenol = 12
volume of ethanol = (12ml/100ml)×750ml = 90ml
Density of Ethanol = 0.789g/ml
Mass of Ethanol = 0.789g/ml × 90ml = 71.01g
Molar mass of ethanol = 46g/mol
No of mole of ethanol = Mass/molar mass = 71.01g/46(g/mol)= 1.5437mole
2) 50g of water = 50ml of water
50ml of water contain 0.48mg of Mercury
Therefore, mercury in 1litre of water = (0.48mg/50ml)×1000ml = 9.6mg
mg/litre = ppm ( parts per million)
1ppm = 1000ppb ( parts per billion )
Therefore,
Mercury concentration in ppm = 9.6ppm
Mercury concentration in ppb = 9600 ppb
3) weight of Compound = 2.819g
Weight of Carbon = 0.102g
weight of X = 2.189g - 0.102g=2.087g
Formula of the compound = CX4
No of C atom = 1
Molar mass of C = 12g/mol
0.102g mol of C combine with 2.087g of X
Therefore, 12g of C combine with (2.087g/0.102g)× 12g =245.53g of X
No of atom of X in CX4 = 4
Therefore , atomic weight of X = 245.53g/4 = 61.38g
4) weight of Iron oxide =4.628g
weight of Fe = 3.348g
Weight of Oxygen = 1.280g
Weight fraction of Fe = 3.348g/4.628g = 0.7234
Weight fraction of Oxygen = 1.280g/4.628g = 0.2766
Therefore,
Formula of Iron Oxide = Fe3O4
5) No of mole = Mass/molar mass
Mass of Al2O3 = 47.51g
Molar mass of Al2O3 = 101.96g/mol
No of mole of Al2O3= 47.51g/101.96(g/mol)
= 0.46596
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