the vapor pressure of ethanol (c2h5oh) at 20 celcius is 44mmHg, and the vapor pr

Question

the vapor pressure of ethanol (c2h5oh) at 20 celcius is 44mmHg, and the vapor pressure of mathanol (ch3oh at the same temperature is 94mmHg. A mixture of 32.9g of methanol and 45.3g of ethanol is prepared and can be assumed to behave as an ideal solution.

calculate the vapor pressure of methanol and ethanol above this solution at 20 celcius.

Calculate the mol fraction of methanol and ethanol in the vapor above this solution at 20 celcius.

vapor pressure?

mathanol__________mmHg

ethanol__________mmHg

calculate mol fraction of both at capor above @ 20 celcius? ethanol______ methanol________

Explanation / Answer

weight of CH3OH taken = 32.9 g
weight of C2H5OH taken = 45.3 g

moeculr mass of CH3OH = 12 + 4 X 1 + 16 = 32 g/mole
molecular mass of C2H5OH = 12X2 + 6 X 1 + 16 = 46 g/mole

no. of moles of CH3OH = 32.9/32 = 1.028125
no. of moles of C2H5OH = 45.3/46 = 0.98478

total no. of moles = 1.028125 + 0.98478 = 2.0129

mole fraction of CH3OH in solution, X(CH3OH) = number of moles of CH3OH/total no. of moles = 1.028125/2.0129 = 0.51

similarly mole fraction of C2H5OH in solution, X(C2H5OH) = 0.978/1.916 = 0.49

vapour pressure of pure CH3OH , Po(CH3OH) = 94 mm
vapour pressure of pure C2H5OH , Po(C2H5OH) = 44 mm

now using raoluts law...

partial pressure of methanol = Po(CH3OH) X X(CH3OH) = 94 X 0.51 = 47.94 mm
partial pressure of ethanol = Po(C2H5OH) X X(C2H5OH) = 44 X 0.49 = 21.56 mm

(b)total pressure = partial pressure of etnalol + partial pressure of methanol = 47.94 + 21.56 = 69.5 mm

mole fraction of methanol in vapour phase = partial pressure of methanol/total pressure = 47.94/69.5 = 0.69

and mole fraction of ethanol in vapour phase = partal pressure of ethanol / total pressure = 21.56/69.5 = 0.31

please check the maths

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