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)
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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
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
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
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
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