Question

a solution contains 18gm of glucose,24gm of acetic acid and 18gm of water .what is the mole fraction of acetic acid.
[Ans:-0.08]

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Answer 1

Answer:answer is 0.26

Explanation:

nC6H12O6= 18/180= 0.1 mol

nCH3COOH = 24g/60g/mol= 0.4 mol

nH2O = 18g/18g/mol = 1mol

Total no of moles= 0.1+0.4+1=1.5

Mole fraction of acetic acid

= 0.4/1.5 = 0.26

Answer 2

Answer:

ANSWER

molality is ratio of (no of moles of solute) and (weight of solvent in kg)

no of moles of solute = grammolecular massgiven mass

no of moles of solute =18018 = 0.1

weight of solvent in kg =1000500 = 0.2

molality=0.20.1

molality=0.5

the answer is 0.5

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