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

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

Please explain further

To calculate the total vapor pressure of a mixture of two liquids, we need to use Raoult's law, which states that the partial pressure of each component in the mixture is proportional to its mole fraction.

First, let's calculate the mole fraction of each component in the mixture:

Methanol (CH3OH):
Molar mass of CH3OH = (12 * 1) + (1 * 4) + (16 * 1) = 32 g/mol
Number of moles of methanol = mass / molar mass = 64 g / 32 g/mol = 2 moles

Ethanol (C2H5OH):
Molar mass of C2H5OH = (12 * 2) + (1 * 6) + (16 * 1) + (1 * 1) = 46 g/mol
Number of moles of ethanol = mass / molar mass = 92 g / 46 g/mol = 2 moles

Now, calculate the mole fraction of each component:
Mole fraction of methanol = moles of methanol / total moles = 2 moles / 4 moles = 0.5
Mole fraction of ethanol = moles of ethanol / total moles = 2 moles / 4 moles = 0.5

Next, use Raoult's law to calculate the partial pressure of each component:

Partial pressure of methanol = mole fraction of methanol * pure vapor pressure of methanol = 0.5 * 90 mmHg = 45 mmHg
Partial pressure of ethanol = mole fraction of ethanol * pure vapor pressure of ethanol = 0.5 * 45 mmHg = 22.5 mmHg

Finally, calculate the total vapor pressure of the mixture by adding the partial pressures together:

Total vapor pressure = Partial pressure of methanol + Partial pressure of ethanol = 45 mmHg + 22.5 mmHg = 67.5 mmHg

Therefore, the total vapor pressure of the mixture of methanol and ethanol at 298K is 67.5 mmHg.