Question

# The vapour pressures of ethanol and methanol are 44.5 mm and 88.7 mm Hg respectively. An ideal solution is formed at the same temperature by mixing 60 g of ethanol with 40 g of methanol. Calculate the total vapour pressure of the solution and the mole fraction of methanol in the vapour.

Solution

## Mol. mass of ethyl alcohol $$=C_2H_5OH=46$$No. of moles of ethyl alcohol $$=\dfrac{60}{46}=1.304$$Mol. mass of methyl alcohol $$=CH_3OH=32$$No. of moles of methyl alcohol $$=\dfrac{40}{32}=1.25$$$$'X_A'$$, mole fraction of ethyl alcohol $$=\dfrac{1.304}{1.304+1.25}=0.5107$$$$'X_B'$$, mole fraction of methyl alcohol $$=\dfrac{1.25}{1.304+1.25}=0.4893$$Partial pressure ethyl alcohol $$=X_A\dot p^0_A=0.5107\times 44.5 = 22.73 mm Hg$$Partial pressure methyl alcohol $$=X_B\dot p^0_B=0.4893\times 88.7 = 73.40 mm Hg$$Total vapour pressure of solution $$=22.73+43.40=66.13 mm Hg$$Mole fraction of methyl alcohol in the vapour$$=\dfrac{\text{Partial pressure of CH}_3OH}{\text{Total vapour pressure}}=\dfrac{43.40}{66.13}=0.6563$$Chemistry

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