Question

5. Given that the molecular weight of ethyl alcohol, CH3CH2OH, is 46, and that of water is 18, how many grams of ethyl alcohol must be mixed with 100 mL of water for the mole fraction (X) of ethyl alcohol to be 0.2?​

Answer 1

Answer : 64 g ethyl alcohol must be mixed

Explanation: we know that density of water is 1 g /ml

therefore, 100 ml have 100 g h2o i.e., 100÷18 moles of h2o

if moles of ethyl alcohol is na and that of h2o is nb , then

na / na+nb = 0.2 (given )

na= 0.2 na + 0.2 nb

na - 0.2 na = 0.2 nb

0.8 na = 0.2 nb

na = 0.2 nb/ 0.8

as we have already calculated that nb i.e., moles of h2o are 100/18

na= (0.2×100) / (0.8×18)

na = 25/18

1 mole of c2h5oh has 46 g of c2h5oh......

by unitary method,

25/18 mole of c2h5oh will have (46×25) / 18 g of c2h5oh

(46×25) / 18 = 64 g...

hope it will help you......

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