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#