sure to aniswer all parts. The vapor pressure of ethanol (C,H,OH) at 20 C is 44
ID: 982316 • Letter: S
Question
sure to aniswer all parts. The vapor pressure of ethanol (C,H,OH) at 20 C is 44 mmi methanol (CH3OH) at the same te and 43.1 g of ethanol is prepared and can be assumed to behave as an ideal solution. Hg, and the vapor pressure of methanol (CH,OlH) at the same temperature is 94 mmilg. A misture of 31.9 g of methanol Calculate the vapor pressure of methanol and ethanol above this solution at 20°C. Be sure to report your answers to the correct number of significant figures. Methanol Ethanol [ ]mmHg [21 ] mmHg Calculate the mole fraction of methanol and ethanol in the vapor above this solution at 20°C. 4 8 21mmHg Ethanol Methanol 0.31 0.69Explanation / Answer
Solution :-
Lets first calculate the moles of the methanol and ethanol
Moles = mass / molar mass
Moles of methanol = 31.9 g / 32.04 g per mol = 0.99563 mol
Moles of ethanol = 43.1 g / 46.068 g per mol = 0.93557 mol
Now lets calculate the new vapor pressure of the methanol and ethanol
Partial pressure of methanol = mole fraction of methanol * vapor pressure of pure methanol
= (0.99563/(0.99563+0.93557)) * 94 mmHg
= 47 mmHg
Partial pressure of ethanol = mole fraction of ethanol * vapor pressure of pure ethanol
= (0.99557/(0.99563+0.93557)) * mmHg
= 22 mmHg
Now lets calculate the total pressure
Total pressure = 22 mmHg + 47 mmHg = 69 mmHg
Now lets calculate the mole fractions of the each methanol and ethanol in the vapor state
Mole fraction of ethanol = partial pressure of methanol / total pressure
= 22 mmHg / 69 mmHg
= 0.319
Mole fration of methanol = partial pressure of ethanol / total pressure
= 47 mmHg / 69 mmHg
= 0.681 mmHg
The values are have very small difference may because of the rounding of the values.
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