Use the amount (mol) of solute and amount (mol) of solvent to calculate the mole fraction. An aqueous solution of ethanol, CH3CH2OH, has a concentration of 0.216 mol/L and has a density of 0.996 g/mL. What are the mass percent and mole fraction of CH3CH2OH in this solution

%m/m = 0.9975%

Xₐ = 0.0039

Explanation:

In order to do this, we need various data. First to all, we need tje molecular mass of the ethanol. this can be obtained in handbooks, or simply taking the atomic weights of carbon (12 g/mol), Hydrogen (1 g/mol) and oxygen (16 g/mol) and summing those values:

MM C₂H₆O = (2*12) + (6*1) + (16*1) = 46 g/mol

Now, there is an expression that is commonly used to determine the molarity of a solution given the mass percent and density, and assuming of course, 1 liter of solution:

M = d * %m/m * 1000 / MM * 100     (1)

From here, we can solve for %m/m:

%m/m = M * MM * 100 / d * 1000

As the problem is not saying the volume of solution, we can easily assume we have 1 liter of solution. Therefore, the %m/m replacing the given data is:

%m/m = 0.216 *46 * 100 / 0.996 * 1000

To get the mole fraction, we first need to get the volume of solvent. From the density, we can get the mass of solution:

m = V * d

m = 1000 * 0.996 = 996 g of solution.

The mass of solute is:

m = 0.216 mol/L * 46 g/mol

m = 9.936 g/L, or simply we have 9.936 g of ethanol in 1 L of solution.

The mass of solvent:

solvent = 996 - 9.936 = 986.064 g

The molecular mass of water, so we can get the moles is 18 g/mol so:

moles water = 986.064 / 18 = 54.78

Finally the mole fraction:

Xₐ = 0.216 / (0.216 + 54.78)

Hope this helps

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