Question

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.

Answer 1

Answer: 158 grams

Step-by-step 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,

`(p^o-p_s)/(p^o)=i* x_2`

where,

`(p^o-p_s)/(p^o)`= relative lowering in vapor pressure

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

`x_2` = mole fraction of solute =
`\frac{\text {moles of solute}}{\text {total moles}}`

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

moles of solute (ethylene glycol) =
`\frac{\text{Given mass}}{\text {Molar mass}}=(x)/(62.07)moles`

moles of solvent (water) =
`\frac{\text{Given mass}}{\text {Molar mass}}=(183g)/(18.02g/mol)=10.2moles`

Total moles = moles of solute (ethylene glycol) + moles of solvent (water) =
`(x)/(62.07)` + 10.2

`x_2` = mole fraction of solute =
`((x)/(62.07))/((x)/(62.07)+10.2)`

`(1.00-0.800)/(1.00)=1* ((x)/(62.07))/((x)/(62.07)+10.2)`

`0.2= ((x)/(62.07))/((x)/(62.07)+10.2)`

`x=158g`

Thus the mass of ethylene glycol should be 158 g