2/10/2016 11:55 PM O 663/1002/10/2016 12:13 PM Print Calaulator Gra Question 2 o
ID: 987288 • Letter: 2
Question
2/10/2016 11:55 PM O 663/1002/10/2016 12:13 PM Print Calaulator Gra Question 2 ol 19 Incorrect In a 61,0g aqueous solution of methanol, CH-O, the mole traction of methanol ls 0.320. What is the mass of each component? Mass of CH40 Number 61 1 Incorrect Start by 1 mol of solution hypothetical solution, and convert them both to grams. Once you know the mass of each component in a find the mass percentage and apply it to the actual solution. the mole fraction to a mass percentage. Assume the moles of each component in that solution, you can Mass of water Number 72.9 g H,o F10 F9 F8Explanation / Answer
Mass of solution = 61 gms
Mole fraction of methanol = 0.320
Mole fraction of water = 1 - 0.320 = 0.680
Molar mass of methanol(CH3OH) = 12 + 4 * 1 + 16 = 32 gn/mol
Molar mass of water = 18 gm/mol
61 = x(0.320 * 32 + 18 * 0.680)
x = 61/22.48 = 2.7135
Number of moles of methanol = 0.320 * 2.7135 = 0.86832 moles
Mass of methanol = number of moles * molar mass = 0.86832 moles * 32 gm/mol = 27.786 gram
Mass of water = 61 - 27.786 = 33.21 gms
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