Question

WHAT MASS OF 1,1 DICHLOROEHTANE MUST BE MIXED WITH 100G OF 1,1 DICHLOROTETRAFLUOROEHTANE TO GIVE A SOLUTION WITH VAPOR PRESSURE 157 TORR AT 25 C

Answer 1

This is an incomplete question.

The complete question is:

1,1-dichlorotetrafluoroethane, CF3CCL2F, has a vapor pressure of 228 torr. What mass of 1,1-dichloroethane must be mixed with 100.0 g of 1,1-dichlorotetrafluoroethane to give a solution with vapor pressure 157 torr at 25 degrees celsius?

Answer: 46.9 g of 1,1 dichloroethane must be fixed with 100 g of 1,1 dichlorotetrafluoroethane to give a solution with vapor pressure of 157 torr at
`25^0C`

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 solute is present in 100 g of solvent

moles of solute (1,1 DICHLOROEHTANE) =
`\frac{\text{Given mass}}{\text {Molar mass}}=(xg)/(98.96g/mol)moles`

moles of solvent (1,1 DICHLOROTETRAFLUOROEHTANE ) =
`\frac{\text{Given mass}}{\text {Molar mass}}=(100g)/(170.92g/mol)=0.58moles`

Total moles = moles of solute + moles of solvent =
`(xg)/(98.96g/mol)+0.58`

`x_2` = mole fraction of solute =
`((xg)/(98.96g/mol))/((xg)/(98.96g/mol)+0.58)`

`(228-157)/(157)=1* ((xg)/(98.96g/mol))/((xg)/(98.96g/mol)+0.58)`

`0.45=1* ((xg)/(98.96g/mol))/((xg)/(98.96g/mol)+0.58)`

`x=46.9g`

Thus 46.9 g of 1,1 dichloroethane must be fixed with 100 g of 1,1 dichlorotetrafluoroethane to give a solution with vapor pressure of 157 torr at
`25^0C`