the ans is wrongwhy Question Anonymous Asked 32 minutes ago

???the ans is wrong?why

Question Anonymous Asked 32 minutes ago asap A corrupt barman attempts to prepare 100.0 em of a drink by mixing 30.0 cm of ethanol with 70.0 cm of water at 25 what volumes should have been mixed in order to arrive at a mixture ofthe same strength but of the required volume (100.0 cm)? The densities of pure water and ethanol are 0.997 and 0.789 glcm at25 respectively. (10%) S4 0.4 0.6 0.8 02 O0 1 solution Anonymous 27 minutes later Strength requird = 30 % Let vol of Ethanol = a Let vol of wtaer = b

Solution

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