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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Answered by
3
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
Answered by
1
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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