Answer :
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 [tex]25^0C[/tex]
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,
[tex]\frac{p^o-p_s}{p^o}=i\times x_2[/tex]
where,
[tex]\frac{p^o-p_s}{p^o}[/tex]= relative lowering in vapor pressure
i = Van'T Hoff factor = 1 (for non electrolytes)
[tex]x_2[/tex] = mole fraction of solute
=[tex]\frac{\text {moles of solute}}{\text {total moles}}[/tex]
Given : x g of solute is present in 100 g of solvent
moles of solute (1,1 DICHLOROEHTANE) = [tex]\frac{\text{Given mass}}{\text {Molar mass}}=\frac{xg}{98.96g/mol}moles[/tex]
moles of solvent (1,1 DICHLOROTETRAFLUOROEHTANE ) = [tex]\frac{\text{Given mass}}{\text {Molar mass}}=\frac{100g}{170.92g/mol}=0.58moles[/tex]
Total moles = moles of solute + moles of solvent = [tex]\frac{xg}{98.96g/mol}+0.58[/tex]
[tex]x_2[/tex] = mole fraction of solute =[tex]\frac{\frac{xg}{98.96g/mol}}{\frac{xg}{98.96g/mol}+0.58}[/tex]
[tex]\frac{228-157}{157}=1\times \frac{\frac{xg}{98.96g/mol}}{\frac{xg}{98.96g/mol}+0.58}[/tex]
[tex]0.45=1\times \frac{\frac{xg}{98.96g/mol}}{\frac{xg}{98.96g/mol}+0.58}[/tex]
[tex]x=46.9g[/tex]
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 [tex]25^0C[/tex]