Calculate the mole fraction of each component of the solution containing 65g of ethanol (C2H5OH) in 350g of water.
χ(A)=n(A)n(A)+n(B)∗100\chi (A)=\frac{n(A)}{n(A) + n(B)} * 100%χ(A)=n(A)+n(B)n(A)∗100 %
χ(H2O)=m(H2O)/M(H2O)m(H2O)/M(H2O) + m(C2H5OH)/M(C2H5OH)∗100\chi (H2O)=\frac{m(H2O)/M(H2O)}{m(H2O)/M(H2O) \ +\ m(C2H5OH)/M(C2H5OH)} * 100%χ(H2O)=m(H2O)/M(H2O) + m(C2H5OH)/M(C2H5OH)m(H2O)/M(H2O)∗100 %
χ(H2O)=350/18350/18 + 65/46∗100\chi(H2O) = \frac{350/18}{350/18 \ +\ 65/46} * 100%χ(H2O)=350/18 + 65/46350/18∗100 % = 93.23%
χ(C2H5OH)=100\chi(C2H5OH) =100χ(C2H5OH)=100 % - χ(H2O)\chi(H2O)χ(H2O) = 100%-93.23% = 6.77 %
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