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Weber State University
Ethanol water mixture has 46% ethanol (weight/ solution weight). Hence, mole fraction of H2O is…. I'll show you two methods that you can use to solve this problem. #color(white)(a)# As you know, a solution's percent concentration by mass tells you the number of grams of solute present for every #"100 g"# of solution. To make the calculations easier, pick a #"100-g"# sample of this solution. Now, you know that the mass of this sample will be equal to the mass of the ethanol, the solute, and the mass of the water, the solvent.
In your case, you will have
You also know that the mole fraction of ethanol, which is defined as the ratio between the number of moles of ethanol and the total number of moles present in the solution, is equal to #0.25#.
At this point, you must use the molar masses of ethanol and of water to express the mole ratio of ethanol in terms of #m_"ethanol"# and #m_"water"#.
This means that you have
Therefore, the mole fraction of ethanol can be rewritten as--for the sake of simplicity, I won't add any units
which is equivalent to
Now all you have to do is to solve this system of two equations with two unknowns. Use equation #color(darkorange)((1))# to write
Plug this into equation #color(darkorange)((2))# to find
This will get you
which results in
Since this represents the mass of ethanol present in #"100 g"# of solution, you can say that the percent concentration by mass of ethanol is
#color(white)(a)# Alternatively, you can start by picking a sample of this solution that contains exactly #1# mole solute and of solvent. This means that you have
Now, you can use the mole fraction of ethanol to say that the number of moles of ethanol present in this sample is equal to
Consequently, you can say that this sample contains #0.75# moles of water. Use the molar masses of the two compounds to convert the number of moles to grams.
The total mass of the solution will be
You can use the known composition of the sample to figure out how many grams of ethanol you'd ge for #"100 g"# of this solution
Once again, you have
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