What is the approximate molar fraction of ethanol c2h5oh in an aqueous solution in which the ethanol concentration is 11.0 molal?
A. 0.0011
B. 0.011
C. 0.170
D. 0.200
E. 0.600
I assume you mean mole fraction, XC2H5OH.
11 m means 11 mols C2H5OH/kg solvent or
11 mols/(11 mol C2H5OH + 1000 gH2O)
Convert 1000 g H2O to mols = about 55 but you need to do it more accurately.
Total mols = 11 + 55 = about 66
XC2H5OH = (11/66) = about 0.167.
To find the approximate molar fraction of ethanol (C2H5OH) in an aqueous solution with an ethanol concentration of 11.0 molal, we can use the formula for molar fraction, which is defined as the ratio of the moles of ethanol to the total moles of all components in the solution.
The molar fraction (X) of ethanol can be calculated as follows:
X = moles of ethanol / total moles of all components
Since we know the ethanol concentration is 11.0 molal, we can assume that there is exactly 11.0 moles of ethanol in 1 kg of water. Thus, the total moles of all components in the solution can be considered as 11.0 moles of ethanol.
Therefore, the molar fraction of ethanol can be calculated as:
X = 11.0 / 11.0 = 1
The molar fraction (X) is a unitless quantity that ranges from 0 to 1, where 0 represents no ethanol presence, and 1 represents pure ethanol. In this case, the molar fraction is 1, indicating that the aqueous solution is entirely ethanol (C2H5OH).
So, the correct answer is not listed among the provided options.
To find the approximate molar fraction of ethanol (C2H5OH) in an aqueous solution with an ethanol concentration of 11.0 molal, we need to use the definition of molarity and molar fraction.
Molality (m) is defined as the number of moles of solute (ethanol) per kilogram of solvent (water). So, in this case, we have 11.0 moles of ethanol per kilogram of water.
Molar fraction (χ) is defined as the ratio of the moles of a component to the total moles of all components in the mixture.
To find the molar fraction of ethanol, we need to calculate the moles of ethanol and the total moles of all components in the mixture.
Since the question states that the ethanol concentration is 11.0 molal, we can assume that the concentration of water is much greater than the concentration of ethanol. Therefore, we can treat water as the solvent and ethanol as the solute.
To calculate the moles of ethanol, we need to know the mass of water. Let's assume that the mass of water is 1000 grams (1 kilogram) for simplicity.
The molar mass of ethanol (C2H5OH) is approximately 46.07 g/mol. Therefore, 11.0 moles of ethanol would have a mass of:
11.0 moles * 46.07 g/mol = 506.77 grams
The molar mass of water (H2O) is approximately 18.02 g/mol. Therefore, 1000 grams of water would have a mass of:
1000 grams / 18.02 g/mol = 55.51 moles
The moles of ethanol divided by the total moles of all components gives us the molar fraction of ethanol:
Molar fraction of ethanol = moles of ethanol / (moles of ethanol + moles of water)
Molar fraction of ethanol = 11.0 moles / (11.0 moles + 55.51 moles) ≈ 0.170
Therefore, the approximate molar fraction of ethanol (C2H5OH) in the aqueous solution is 0.170.
The answer is C. 0.170.