At a certain temperature the vapor pressure of pure water is measured to be 212. mmhg. suppose a solution is prepared by mixing 121. g of water and 59.6 g of methanol (ch3oh) calculate the partial pressure of water vapor above this solution. be sure your answer has the correct number of significant digits note for advanced students: you may assume the solution is ideal mmhg

The partial pressure of water vapor 166 mmHg

Explanation:

Step 1: Data given

Vapor Pressure = 212 mmHg =  212 /760 = 0.278947 atm

Mass of water = 121 grams

Mass of methanol = 59.6 grams

Molar mass of water = 18.02 g/mol

Molar mass of methanol = 32.04 g/mol

Step 3: Calculate moles of water

Moles water = mass water / molar mass water

Moles water = 121 grams / 18.02 g/mol

Moles water = 6.71 moles

Step 4: Calculate moles methanol

Moles methanol = 59.5 grams / 32.04 g/mol

Moles methanol = 1.86 moles

Step 5: Calculate total moles

Total moles = moles water + moles methanol

Total moles = 6.71 moles + 1.86 moles = 8.57 moles

Step 6: Calculate mol fraction of water

Mol fraction of water = 6.71 moles / 8.57 moles

Mol fraction of water = 0.782

Step 7: Calculate partial pressure of water vapor

Partial pressure H2O = mol fraction water * vapor pressure

Partial pressure H2O = 0.782 * 212 mmHg = 165.8 mmHg

The partial pressure of water vapor  is 165.8 ≈ 166 mmHg

it is b that is the answer.

0.88m hope i am correct : )