Question

If the initial concentration of ethanol was 13.5% v/v. What is the mole fraction of ethanol and water? (n water = 1.3348) and (n ethanol = 1.3639). Assuming (50 mL solution).

Answer 1

First find total no. of mole of solution

13.5% v/v means 13.5 ml Ethanol is present in 100 ml of solution,rest 86.5 ml water present in 100 ml of solution

in 50 ml of solution volume of ethanol is 13.5x50/100=6.75% and volume of water is 86.5x50/100=43.25%

No.of mole of ethanol= volume at NTP in litre /standard molar volume(22.4L)

volume of ethanol in litre=6.75/1000=0.00675 L

no.of mole of ethanol= 0.00675/22.4=0.000301

volume of water in litre=43.2/1000=0.0432 L

No.of mole of water=0.0432/18=0.0024

total no. of mole of solution(ethanol +water)=0.000301+0.0024=0.002702

Mole fraction = Given mole of solute or solvent/total no. of mole of solution

in question given no, of mole of ethanol is 1.3639 and mo.of mole of water is 1.3348

mole fraction of ethanol=1.3639/0.002702=504.774

mole fraction of water=1.3348/0.002702=494.004