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3 years ago

A solution is prepared by mixing 25.0 g H2O and 25.0 g C2H5OH. Determine the mole fractions of each substance.

Chemistry

2 answers:

Veronika [31]3 years ago

8 0

Answer:

Mole fraction H₂O → 0.72

Mole fraction C₂H₅OH → 0.28

Explanation:

By the mass of the two elements in the solution, we determine the moles of each:

25 g . 1 mol/ 18g = 1.39 moles of water (solute)

25 g . 1 mol / 46 g = 0.543 moles of ethanol (solvent)

Mole fraction solute = Moles of solute / Total moles

Mole fraction solvent = Moles of solvent / Total moles

Total moles = Moles of solute + Moles of solvent

1.39 moles of solute + 0.543 moles of solvent = 1.933 moles → Total moles

Mole fraction H₂O = 1.39 / 1.933 → 0.72

Mole fraction C₂H₅OH= 0.543 / 1.933 → 0.28

Remember that sum of mole fractions = 1

MArishka [77]3 years ago

6 0

Answer:

Mol fraction H2O = 0.72

Mol fraction C2H5OH = 0.28 moles

Explanation:

Step 1: Data given

Mass of H2O = 25.0 grams

Molar mass H2O = 18.02 g/mol

Mass of C2H5OH = 25.0 grams

Molar mass C2H5OH = 46.07 g/mol

Step 2: Calculate moles

Moles = mass / molar mass

Moles H2O = 25.0 grams / 18.02 g/mol

Moles H2O = 1.39 moles

Moles C2H5OH = 25.0 grams / 46.07 g/mol

Moles C2H5OH = 0.54 moles

Step 3: Calculate total moles

Total moles = 1.39 moles + 0.54 moles

Total moles = 1.93 moles

Step 4: Calculate mol fractions

Mol fraction = moles / total moles

Mol fraction H2O = 1.39 moles / 1.93 moles

Mol fraction H2O = 0.72

Mol fraction C2H5OH = 0.54 moles / 1.93 moles

Mol fraction C2H5OH = 0.28 moles

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Let's combine the Ideal Gases Law with density to get the molecular formula for the unknown gas.

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760 Torr are 1 atm

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0.488 atm . 1L = 1.06g/MM . 0.082 . 304K

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3 years ago

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Answer:

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Explanation:

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Since all of the answer choices are given in units of Liters, it will be convenient to use a value for R that contains "Liters" in its units: $R=0.0821\frac{L\cdot atm}{mol\cdot K}$

Since the conditions are stated to be STP, we must remember that STP is Standard Temperature Pressure, which means $T=273.15K$ and $P=1atm$

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...substituting the known values, and simplifying...

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