Question

120g acetic acid ch3cooh is present in 360g of water, find out mole fraction of every constituent

Answer 1

HEY MATE ❤❤♥♥♥♥♣♣♣♣♣♠♠♠♠❤❤❤❤❤♥♥♥♥♥♥♥♥♥♥♥ answer here's ❤❤♥♥❤❤♥♥❤❤♥♥Moles of water in mixture=12/18=2/3

moles of acetic acid in mixture = 108/60 = 18/10

moles of ethanol in mixture= 92/46 = 2

mole fraction of water = (2/3)/(2/3+1.8+2) = (2/3)/[(2+5.4+6)/3] =2/13.4=20/134 = 10/67

therefore 10/67 is the mole fraction of water in mixture.mole fraction= moles of solute/moles of solution

To find out the more fraction more data must be present including information about solvent and mass of acetic acid present in it.

Molar mass of acetic acid(CH3COOH) can be given by:

carbon+oxygen +hydrogen

=12*2+16*2+4*1

hope this helps:p

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