Calculate molarity of solution of ethanol in water in which the mole fraction of ethanol is 0.040

Calculate molarity of solution of ethanol in water in which the mole f

1.	Calculate molarity of solution of ethanol in water in which the mole fraction of ethanol is 0.040
Answer» Let the mole fraction of ethanol (C2H5OH)\xa0be represented by x,Then the following expression represents its mole fraction in ethanol solution-\xa0xC2h5OH =\xa0{tex}\\frac{\\mathrm{n}\\left(\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}\\right)}{\\mathrm{n}\\left(\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}\\right)+\\mathrm{n}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)}{/tex}(where x\xa0C2H5OHrepresents mole fraction of ethanol)= 0.040 (given)....(i);where "n\'\' (represents number of moles of the two referred liquids in ethanol solution).The following steps are used for\xa0calculations,The aim is to find number of moles of ethanol in 1 L of the solution.Step 1 /- No. of moles in 1 L of water=\xa0{tex}\\frac{1000 \\mathrm \\ {g}}{18 \\mathrm \\ {g} \\mathrm \\ {mol}^{-1}}{/tex}= 55.55 molesStep 2 /-. Substituting n (H2O)= 55.55 in equation (i), we getx (\xa0C2H5OH)=\xa0{tex}\\frac{\\mathrm{n}\\left(\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}\\right)}{\\mathrm{n}\\left(\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}\\right)+55.55}{/tex}= 0.040......(given )solving for n (C2H5OH) we get,0.96 n ( C2\xa0H5\xa0OH )\xa0= 55.55\xa0{tex}\\times{/tex} 0.040{tex}\\therefore{/tex}\xa0n (C2H5OH)= 2.31 molsSo 2.31 moles of C2\xa0H5\xa0OH are present per litre of given ethanol solution.\xa0Hence,the\xa0molarity of the solution of ethanol is\xa02.31 M.