A 2.9 m solution of methanol in water has a density of 0.984gm/ml what are (a)mass percent (b)molarity and (c)mole percentof methanol in  the solution.

Asked by whitestone7587 | 16th Jul, 2014, 05:43: PM

Expert Answer:

moles CH3OH = 2.90 m = 2.90 moles/kg

Mass of CH3OH = 2.90 mol x 32.042 g / mol = 92.9 g

mass of water = 1000 g

mass of solution = 1000+ 92.9 = 1092.9 g

Volume of solution = 1092.9/0.984 = 1110 mL= 1.11 L

moles of water = 1000 g / 18.02 g/mol = 55.5 moles

% by mass of CH3OH = 92.9 x 100 / 1092.2 = 8.44 %

M= 2.90 / 1.11 = 2.61 moles / L

mole percent of CH3OH= 2.90 x 100 / (2.90+ 55.5) = 2.90 x 100 / 58.4 = 4.96 %

Answered by Prachi Sawant | 17th Jul, 2014, 10:44: AM

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