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Mole fraction of C2H5OH = No. of moles of C2H5OH/No. of moles of solution
nC2H5OH = n(C2H5OH)/(C2H5OH)+n(H2O) = 0.040 (Given) … 1
We have to find the number of moles of ethanol in 1L of the solution but the solution is dilute. Therefor, water is approx. 1L.
No. of moles in 1L of water = 1000g/18g mol-1 = 55.55 moles
Substituting n(H2O) = 55.55 in equation 1
n(C2H5OH)/(C2H5OH) + 55.55 = 0.040
⇒ 0.96n(C2H5OH) = 55.55 × 0.040
⇒ n(C2H5OH) = 2.31 mol
Hence, molarity of the solution = 2.31M
nC2H5OH = n(C2H5OH)/(C2H5OH)+n(H2O) = 0.040 (Given) … 1
We have to find the number of moles of ethanol in 1L of the solution but the solution is dilute. Therefor, water is approx. 1L.
No. of moles in 1L of water = 1000g/18g mol-1 = 55.55 moles
Substituting n(H2O) = 55.55 in equation 1
n(C2H5OH)/(C2H5OH) + 55.55 = 0.040
⇒ 0.96n(C2H5OH) = 55.55 × 0.040
⇒ n(C2H5OH) = 2.31 mol
Hence, molarity of the solution = 2.31M
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