What mass of ethylene glycol, when mixed with 183 g H2O, will reduce the equilibrium vapor pressure of H2O from 1.00 atm to 0.800 atm at 100 °C? The molar masses of water and ethylene glycol are 18.02 g/mol and 62.07 g/mol, respectively. Assume ideal behavior for the solution.

158 grams

Explanation:

As the relative lowering of vapor pressure is directly proportional to the amount of dissolved solute.

The formula for relative lowering of vapor pressure will be,

where,

= relative lowering in vapor pressure

i = Van'T Hoff factor = 1 (for non electrolytes)

= mole fraction of solute  =

Given : x g of ethylene glycol is present in 183 g of water

moles of solute (ethylene glycol) =

moles of solvent (water) =

Total moles = moles of solute (ethylene glycol)  + moles of solvent (water) = + 10.2

= mole fraction of solute =

Thus the mass of ethylene glycol should be 158 g

b is the correct answer

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