just c ..and please give more details. also we don\'t know what the percentage o

Question

just c ..and please give more details.

also we don't know what the percentage of ethyl acetate in the sample so how we can get the mass of the sample because they give percentage of mixture..are the percentage of ethyl acetate the mixture = the percentage of ethyl acetate in the sample?

A mixture is 10.0 mole% ethyl alcohol (GHOH), 75.0 moles ethyl acetate (C and 15.0 mole% acetic acid (CHeogH), la 050. 01 (a) Calculate the mass fractions of each compouned. (b) What is the average molecular weight of the mixture? (c) What would be the mass (Cg) of a sample containing t /6) [/21 25.0 kmol of ethyl acetate?

Explanation / Answer

a)

mass fraction of x = mass of x / total mass * 100%

for ethyl alcohol

Assume 100 mol of mix

mass of ethyl alchole = mol*MW = 10*46 = 460 g

mass of ethyl acetate = mol*MW = 75*88.1051 = 6607.882 g

mass of acetic acid = mol*MW = 15*60 = 900 g

% ethyl alcohol = 460 / (460+6607.882 +900 ) * 100 = 5.773%

% ethyl acetate = 6607.882 / (460+6607.882 +900 ) * 100 = 82.93%

%acetic acid = 900 / (460+6607.882 +900 ) * 100 = 11.295 %

b)

Average molecular weigh of mix

MW avg = x1*MW1 + x2*MW + x3*M3

MW avg = 0.10 * 46 + 0.75 * 75 + 0.15*60 = 69.85 g/mol

c)

if 25 kmol

75% --> 25 kmol

100% --> 25/0.75 = 33.33 kmol

Total mass = kmol*MW=  33.33 *69.85 = 2328.10 kg of mix

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