The vapor pressure of water at 20°C is 17.54 torr. What is the vapor pressure (in torr)of a solution containing 143.8g of dextrose (C6H12O6 ) and 146.5 g of water at ?

Answer 1

I see a question mark at the end of at..? Maybe it means "what is the vapor pressure of a solution at ... 20ºC?

That being the case, we would want to consider Raoult's Law. Whenever you add a solute to a pure solvent, the vapor pressure of that pure solvent will decrease. The amount the vapor pressure decreases will depend on the amount of solute you add (and hence on the amount of pure solvent remains). You determine this by finding the mole fraction of the pure solvent.

Mole fraction = moles of substance / total moles

For the solvent, H2O, we have 146.5 g H2O x 1 mol H2O/18 g = 8.14 moles H2O

For the solute, C6H12O6, we have 143.8 g x 1 mol / 180 g = 0.799 moles C6H12O6

Total moles = 8.94 moles

Mole fraction H2O = 8.14 mol/8.94 mol = 0.911

So, essentially, the H2O is only 91.1% pure and so we find the vapor pressure of the solution as follows:

0.911 x 17.54 torr = 16.0 torr

You may see this above equation written more scientifically as Raoult's Law as ...

P_x = X_xP_xº

P = partial pressure of solvent

X = mole fraction of solvent

Pº = vapor pressure of pure solvent