This is a Raoult's law problem.
Raoult's law states that the vapour pressure of a solution of a non-volatile solute is equal to the vapour pressure of the pure solvent at that temperature multiplied by its mole fraction.
In symbols,
#p = chi_1p_1^@#
where
#chi#the mole fraction of the solvent
#p# and #p_1^@# are the vapour pressures of the solution and of the pure solvent
The vapour pressure lowering is given by
#Δp = p_1^@ - p_1 = p_1^@ - chi_1p_1^@ = p_1^@(1 - chi_1)#
#1 - chi_1 = chi_2#, so,
#Δp = chi_2p_1^@#
where #chi_2# is the mole fraction of the solute.
Step 1. Calculate the moles of each substance
#"Moles of glucose" = n_2 = color(red)(cancel(color(black)(18 "g glucose"))) × "1 mol glucose"/(180.16 color(red)(cancel(color(black)("g glucose")))) = "0.0999 mol glucose"#
#"Moles of water" = n_1 = 178.2 color(red)(cancel(color(black)("g H"_2"O"))) × ("1 mol H"_2"O")/(18.02 color(red)(cancel(color(black)("g H"_2"O")))) = "9.889 mol H"_2"O"#
#"Total moles" = n_2 + n_1 = "(0.0999 + 9.889) mol" = "9.989 mol"#
Step 2. Calculate the mole fraction of glucose
#chi_2 = n_2/(n_1 + n_2) = (0.0999 color(red)(cancel(color(black)("mol"))))/(9.989 color(red)(cancel(color(black)("mol")))) = 0.01000#
Step 3. Calculate the vapour pressure lowering
At 373 K, #"p1^@ = "101.3 kPa"#
#Δp = chi_2p_1^@ = 0.0100 × "101.3 kPa" = "10.13 kPa"#
Step 4. Calculate the vapour pressure
#p_1 = p_1^@ - Δp = "(101.3 - 10.13) kPa = 91.2 kPa"#
