Mole fraction of ethanol in ethanol water mixture is 0.254 .hence percentage concentration of ethanol by weight in mixture is.
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Answer:
percentage concentration of ethanol by weight = 46 %
Explanation:
Given
Mole fraction of ethanol = X2 = 0.254
Mole fraction of water = X1 =1-0.254
percentage concentration of ethanol by weight = ??
Solution
molar weight of ethanol = mol.wt2= 46 g/mol
molar weight of water = mol.wt1= 18 g/mol
we can write
n1 = W1/18 ----------1
n2 = W2/46------------2
SIMILARLY
X1 = n1/ n1 + n2--------3
X2 = n2/ n1 + n2---------4
from eq 3 and 4 we get
X1 / X2 = n1 / n2
n1 / n2 = 0.746/ 0.254
from eq 1 and 2
W1/18 divided by W2/46 = 3/1
W1 X46/ W2 X18 = 3/1
W1/W2 = 3 X 18 / 46
W1/W2 = 1.17
percentage concentration of ethanol
percentage concentration of ethanol W2 = (W2 / W1 + W2) 100
% W2 = (W2 / W1 divided by 1 + W2/W1 ) 100
putting values
% W2 = (1/1.17 divided by 1 + 1/1.17 ) 100
% W2 = 0.46 X 100
% W2 = 46 %
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