1.

A solution contains 25% water, 25% ethanol and 50% acetic acid by mass. Calculate the mole fraction of each component.

Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :Suppose, we have 100 g of the <a href="https://interviewquestions.tuteehub.com/tag/given-473447" style="font-weight:bold;" target="_blank" title="Click to know more about GIVEN">GIVEN</a> solution. Then, <br/> mass of <a href="https://interviewquestions.tuteehub.com/tag/water-1449333" style="font-weight:bold;" target="_blank" title="Click to know more about WATER">WATER</a> = 25g mass of ethanol = 25 g and mass of acetic acid = 50 g. <br/> The molecular <a href="https://interviewquestions.tuteehub.com/tag/masses-1088822" style="font-weight:bold;" target="_blank" title="Click to know more about MASSES">MASSES</a> of water, ethanol and acetic acid are 18, 46 and 60 <a href="https://interviewquestions.tuteehub.com/tag/respectively-1186938" style="font-weight:bold;" target="_blank" title="Click to know more about RESPECTIVELY">RESPECTIVELY</a>. Therefore,<br/> number of moles of water `(n_(1)) = 25/18 = 1.39` <br/> number of moles of ethanol `(n_(2)) = 25/46 = 0.54` <br/> and number of moles of acetic acid `(n_(3)) = 50/60 = 0.83` <br/> Total number of moles in solution = `n_(1) + n_(2) + n_(3) = 1.39 + 0.54 + 0.83 = 2.76`<br/> Mole fraction of water = `n_(1)/(n_(1) + n_(2)+ n_(3))` <br/> `=1.39/2.76 = 0.503` <br/> Mole fraction of ethanol `=n_(2)/(n_(1) + n_(2) + n_(3))` <br/> `=0.54/2.76 = 0.196` <br/> Mole fraction of acetic acid = `n_(3)/(n_(1) + n_(2) + n_(3)) = 0.83/2.76 = 0.301`</body></html>


Discussion

No Comment Found

Related InterviewSolutions