We're asked to find the mole fraction of #"KCl"# given that is present in a mass percentage of #26.3%#.
This means that in a #100#-#"g"# sample of solution, there are #26.3# #"g KCl"# and #73.7# #"g H"_2"O"#.
Let's convert these mass values to grams using the molar mass of each substance:
#26.3cancel("g KCl")((1color(white)(l)"mol KCl")/(74.55cancel("g KCl"))) = 0.353# #"mol KCl"#
#73.7cancel("g H"_2"O")((1color(white)(l)"mol H"_2"O")/(18.02cancel("g H"_2"O"))) = 4.09# #"mol H"_2"O"#
The mole fraction of #"KCl"# is
#chi_"KCl" = "mol KCl"/"total moles" = (0.353color(white)(l)"mol KCl")/(0.353color(white)(l)"mol KCl" + 4.09color(white)(l)"mol H"_2"O")#
#= color(red)(0.0794#
