3.A corrupt barman attempts to prepare 100.0 cm2 of a drink by mixing 30.0 cm\'

Question

3.A corrupt barman attempts to prepare 100.0 cm2 of a drink by mixing 30.0 cm' of ethanol with 70.0 cm'of water at 25 °C. What volumes should have been mixed in order to arrive at a mixture of the same strengt but of the required volume (100.0 cm')? The densities of pure water and ethanol are 0.997 and 0.789 g/cm at 25°C, respectively. (10%) 58 16 56 Water 16 54 0.6 0.8 Mole fraction of ethanol,x(CH,O0 .Cyclobutadiene, CaHs, is a four-carbon atom ring. Write the secular determinantal ea molecular orbitals of this planar molecule. Find the Hückel molecular orbital energies and show the HOM and LUMO Predict the total electronic energy ofthis compound. Is there extra electron stabilization (as in butadiene or benzene) in this molecule? (10%) at are the three types of reaction common to all chain mechanisms? H2+ Bry reaction mechanism involves free radical carriers mlu

Explanation / Answer

Original solution:

Volume of ethanol mixed = 30 cm3

Mass of ethanol mixed = 30 * 0.789 g = 23.67 g

Moles of ethanol mixed = 23.67 g / 46.07 g/mol

= 0.514

Volume of water mixed = 70 cm3

Mass of water mixed = 70 * 0.997 g = 69.79 g

Moles of water mixed = 69.79 g / 18.015 g/mol

= 3.87

Mole fraction of ethanol = 0.514 / (0.514 + 3.87)

= 0.117

Mole fraction of water = 1 – 0.117

= 0.883

From the graph,

Partial volume of ethanol at 0.117 mole fraction = 53 cm3/mol

Partial volume of water at 0.883 mole fraction = 15 cm3/mol

Let x moles of ethanol are to be added.

For desired strength, moles of water added = 0.883/0.117 x = 7.54 x

Total volume = 100 cm3

Total volume = 53 x + 15 * 7.54 x = 100

x = 0.6

Ethanol moles to be added, x = 0.6 moles

Water moles to be added = 7.54 x = 4.54

Mass of ethanol to be added = 0.6 * 46.07 = 27.736 g

Mass of water to be added = 4.54 * 18.015 = 81.78 g

Volume of ethanol to be added = Mass / density

= 27.736 / 0.789 = 35.15 cm3 ~ 35 cm3

Volume of water to be added = Mass / density

= 81.78 / 0.997 = 82.02 cm3 ~ 82 cm3

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