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
