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Q. An ideal solution was obtained by mixing methanol and ethanol. If the partial vapour pressure of methanol and ethanol are $2.619 \,kPa$ and $4.556 \,kPa$ respectively; the composition of vapour (in terms of mole fraction) will be

Solutions

Solution:

$ P_{ Total }=2.619+4.556=7.175 kPa$

Mole fraction in vapour phase $=\frac{\text { Partial pressure }}{\text { Total pressure }}$

$X_{ MeOH }=\frac{2.619}{7.175}=0.365 $

$\Rightarrow X_{ EtOH }=1-0.365=0.635$