Asked by ruth

calculate the total vapour pressure of mixture of methanol(64g) and ethanol(92g) at 298K given that the pure vapour pressure of methanol is 90mmHg and ethanol is 45mmHg (C=12, H=1, O=16)
Answered by Anonymous
geez - pick a name and stay with it!
Answered by Doc48
VP(MtOH) = 90mmHg m(MtOH) = 64g/32g/mole = 2 mole
VP(EtOH) = 45mmHg m(EtOH) = 92g/46g/mole = 2 mole
X(MtOH) = X(EtOH) = 2/(2 + 2) = 0.50
VP(MtOH) in mix = X(MtOH)∙VP(MtOH) = 0.50(90mmHg) = 45mmHg
VP(EtOH) in mix = X(ETOH)∙VP(EtOH) = 0.50(45mmHg) = 22.5mmHg
TTL Pressure of mix = 45.0mmHg + 22.5mmHg = 67.5mmHg
Answered by Doc48
Correction...
Total Pressure = 90 + 45 = 135mmHg
VP(MtOH) in mix = X(MtOH)∙VP(MtOH) = 0.50(135mmHg) = 67.5mmHg
VP(EtOH) in mix = X(ETOH)∙VP(EtOH) = 0.50(135mmHg) = 67.5 mmHg
TTL Pressure = 135mmHg
Answered by Bajo
Please explain further
There are no AI answers yet. The ability to request AI answers is coming soon!