Vapor Pressure Problem Please Help? Chemistry 2?

Wouldn't it be negative?

#DeltaP_A = -"0.1238 kPa"#

So then, what's the new vapor pressure?

The vapor pressure of an ideal solution containing a nonvolatile electrolyte is given by Raoult's law.

#P_A = chi_(A(l)) P_A^"*"#

where #A# is solvent, #chi_(A(l))# is its mol fraction in the liquid phase, #P_A# is its partial pressure above itself, and #"*"# indicates pure solvent.

Since you have over #"10 mols"# of water, clearly that's your solvent. But we shouldn't have to know its mol fraction to know how the vapor pressure changes.

By definition, changes are final minus initial. Our final state is the solution, and the initial state is the pure solvent...

#DeltaP_A = P_A - P_A^"*"#

#= chi_(A(l))P_A^"*" - P_A^"*"#

#= (chi_(A(l))-1)P_A^"*"#

Since

#chi_(A(l)) + chi_(B(l)) = 1#,

where #B# is solute,

#color(blue)(DeltaP_A) = (1 - chi_(B(l)) -1)P_A^"*"#

#= color(blue)(-chi_(B(l))P_A^"*")#

Mol fractions are nonnegative, and so are pressures.

Thus, the change in vapor pressure due to adding ANY nonvolatile solute at all must be negative.

The mols of fructose are:

#73.40 cancel"g Fruc" xx ("1 mol")/(180.156 cancel"g Fruc") = "0.4074 mols"#

The mols of water are:

#180.5 cancel"g water" xx ("1 mol")/(18.015 cancel"g water") = "10.02 mols"#

Therefore, the mol fraction of fructose is:

#chi_(B(l)) = n_(B(l))/(n_(A(l)) + n_(B(l)))#

#= ("0.4074 mols")/("0.4074 mols" + "10.02 mols")#

#= 0.03907#

and so, the CHANGE in vapor pressure of water is:

#color(blue)(DeltaP_A) = -0.03907 cdot "3.1690 kPa"#

#= color(blue)(-"0.1238 kPa")#

and the DECREASE in vapor pressure of water is:

#color(blue)(|DeltaP_A|) = |-0.03907 cdot "3.1690 kPa"|#

#= color(blue)(+"0.1238 kPa")#

Why did I bother to distinguish between these?