# Chemistry Question about moles/mass percent/molality?

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A solution is made containing 24.5 g phenol (C6H5OH) in 480 g ethanol (CH3CH2OH).

Calculate the mole fraction of phenol.

Calculate the mass percent of phenol.

Calculate the molality of phenol.

Thanks in advance, I can’t figure these out!

• Molar mass of phenol = 94 g/mol

Moles of phenol = 24.5 g / (94 g/mol) = 0.2606 mol

Molar mass of ethanol = 46 g/mol

Moles of ethanoll = 480 g / (46 g/mol) =10.435 mol

——————–

Mole fraction of phenol

X_phenol = Moles phenol / (moles phenol+moles ethanol)

X_phenol = 0.2606 / (0.2606 + 10.435)

X_phenol = 0.02437

——————–

Mass percent of phenol

%phenol = mass phenol / (mass phenol+mass ethanol)*100

% phenol = 24.5 / (24.5 + 480) * 100

% phenol = 4.856 %

——————–

Molality of phenol

m_phenol =moles phenol / (kg of ethanol)

m_phenol = 0.2606 / 0.480

m_phenol = 0.5429 mol / kg

Bye.

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• 24.5 g phenol = 0.260 mole (24.5 g / 94.114 g/mol = 0.260 mol)

480g ethanol = 10.42 mole (480g / 46.07 g/mol = 10.42 mol)

Total moles = 10.68 (0.26 + 10.42 = 10.68)

Mole fraction phenol = 0.024 (0.26/10.68 = 0.024)

Total mass = 504.5 g (24.5 + 480g = 504.5g)

Mass percent of phenol = (24.5 / 504.5) x 100 = 4.86%

Molality = moles solute per 1000 g solvent

Molality = 0.26 / 0.480 = 0.542 molal

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to find mass from moles, multiply the number of moles of the substance by the molar mass (which the the sum of all of the molar masses of the elements in the substance) 1. 0.436 moles of ammonium chloride = 0.436 mol (53.49 g/mol) = 23.32164. then you have to round it to 3 sig figs because .436 has 3 sig figs. so the answer is 23.3 g of ammonium chloride. 2. 2.360 moles of lead (II) oxide = 2.360 mol (223.20 g/mol) = 526.8 g of lead (II) oxide 3. 0.031 moles of aluminum iodide = 0.031 mol (407.695 g/mol) = 13 g of aluminum iodide 4. 1.077 moles of magnesium phosphate = 1.077 mol (262.85 g/mol) = 283.1 g of magnesium phosphate 5. 0.50 moles of calcium nitrate = 0.50 mol (164 g/mol) = 82 g of calcium nitrate to find the moles from mass, divide the number of grams of a substance by the molar mass of the substance 6. 23.5 g of sodium chloride = 23.5 g/58.44 g/mol = 0.4021218. Round to 3 sig figs and you get .402 mol of sodium chloride 7. 0.778 g of sodium cyanide = 0.778 g/49.01 g/mol = 0.016 mol of sodium cyanide 8. 0.250 g of water = 0.250 g/18 g/mol = 4.5 mol of water 9. 169.45 g of calcium acetate = 169.45 g/158.138 g/mol = 1.0715 mol of calcium acetate 10. 79.9 g of potassium permanganate = 79.9 g/158.03 g/mol = 0.506 mol of potassium permanganate

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