The formula for vapour pressure lowering #Δp# is
#color(blue)(barul|stackrel(" ")(Δp = χ_2p_1^@)|)#
where
#χ_2 =# the mole fraction of the solute
#p_1^@ = # the vapour pressure of the pure solvent
However, we must remember that vapour pressure lowering is a colligative property: it depends on the total number of particles in the solution.
Hence, when we are dealing with strong electrolytes, we must add the
van't Hoff #i# factor to the equation.
The equation then becomes
#color(blue)(barul|stackrel(" ")(Δp = iχ_2p_1^@)|)#
Since the solvent is the same in every case, all we need to calculate is the value
of #iχ_2#.
Assume we have 1000 g of each solution.
a) 10 % #"MgCl"_2#
#"Mass of MgCl"_2 = "100 g"#
#"Moles of MgCl"_2 = 100 color(red)(cancel(color(black)("g MgCl"_2))) × "1 mol MgCl"_2/(95.21 color(red)(cancel(color(black)("g MgCl"_2)))) = "1.05 mol MgCl"_2#
#"Mass of water = 900 g"#
#"Moles of water" = 900 color(red)(cancel(color(black)("g water"))) × "1 mol water"/(18.02 color(red)(cancel(color(black)("g water")))) = "49.9 mol water"#
#χ_2 = n_2/(n_2 + n_1) = (1.05 color(red)(cancel(color(black)("mol"))))/((1.05 + 49.9) color(red)(cancel(color(black)("mol")))) = 1.05/51.0 = 0.0206#
#iχ_2 = 3 × 0.0206 = 0.0618#
b) 10 % #"CaCl"_2#
#"Mass of CaCl"_2 = "100 g"#
#"Moles of CaCl"_2 = 100 color(red)(cancel(color(black)("g CaCl"_2))) × "1 mol CaCl"_2/(110.98 color(red)(cancel(color(black)("g CaCl"_2)))) = "0.901 mol CaCl"_2#
#χ_2 = n_2/(n_2 + n_1) = (0.901 color(red)(cancel(color(black)("mol"))))/((0.901 + 49.9) color(red)(cancel(color(black)("mol")))) = 0.901/50.8 = 0.0177#
#iχ_2 = 3 × 0.0177 = 0.0532#
c) 10 % #"KCl"#
#"Mass of KCl = 100 g"#
#"Moles of KCl" = 100 color(red)(cancel(color(black)("g KCl"))) × "1 mol KCl"/(74.55 color(red)(cancel(color(black)("g KCl")))) = "1.34 mol KCl"#
#χ_2 = n_2/(n_2 + n_1) = (1.34 color(red)(cancel(color(black)("mol"))))/((1.34 + 49.9) color(red)(cancel(color(black)("mol")))) = 0.901/51.3 = 0.0262#
#iχ_2 = 2 × 0.0262 = 0.0523#
d) 10 % #"NaCl"#
#"Mass of NaCl = 100 g"#
#"Moles of NaCl" = 100 color(red)(cancel(color(black)("g NaCl"))) × "1 mol NaCl"/(58.44 color(red)(cancel(color(black)("g NaCl")))) = "1.71 mol NaCl"#
#χ_2 = n_2/(n_2 + n_1) = (1.71 color(red)(cancel(color(black)("mol"))))/((1.71 + 49.9) color(red)(cancel(color(black)("mol")))) = 1.71/51.7 = 0.0262#
#iχ_2 = 2 × 0.0331 = 0.0663#
The solution of 10 % #"NaCl"# has the greatest value of #iχ_2#, so it has the greatest vapor pressure lowering and the lowest vapour pressure.
