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 $25^0C$

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,

$\frac{p^o-p_s}{p^o}=i\times x_2$

where,

$\frac{p^o-p_s}{p^o}$= 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) = $\frac{\text{Given mass}}{\text {Molar mass}}=\frac{xg}{98.96g/mol}moles$

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

Total moles = moles of solute  + moles of solvent = $\frac{xg}{98.96g/mol}+0.58$

$x_2$ = mole fraction of solute   =$\frac{\frac{xg}{98.96g/mol}}{\frac{xg}{98.96g/mol}+0.58}$

$\frac{228-157}{157}=1\times \frac{\frac{xg}{98.96g/mol}}{\frac{xg}{98.96g/mol}+0.58}$

$0.45=1\times \frac{\frac{xg}{98.96g/mol}}{\frac{xg}{98.96g/mol}+0.58}$

$x=46.9g$

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 $25^0C$