This question is impossible to answer without making some assumptions.
I am going to assume that we are dealing with a "w/w" (weight by weight) percentage solution (for more information on percentage solutions please see this blog post: Calculating Percentage Solutions). The reason I am working on "w/w" is because if this is a "w/v" solution I would have 26.3 g of KCl in a final volume of 100 ml, and as we will see, I need to know the exact amount of water added. The difference between a "w/w" and "w/v" solution is most probably negligible, but science is all about being accurate.
The mole fraction is the number of moles of something divided by the number of total moles present.
"mole fraction" = "moles" / "total moles present"
First, let us work out the number of moles of KCl present. A 26.3% "w/w" solution would contain 26.3 g of KCl per 100 g of solution.
The molecular weight of KCl is: 74.5513 g/mol
Hence, 26.3 g of KCl is 26.3 / 74.5513 = 0.353 moles.
A "w/w" percentage solution has a final weight of 100 g. We have 26.3 g of KCl, so the water must weigh 100 - 26.3 = 73.7 g.
The molecular weight of water is: 18.01528 g/mol.
Hence we have, 73.7 / 18.01528 = 4.091 moles.
Therefore we have:
"mole fraction" = 0.353 / ("("0.353 + 4.091")")
"mole fraction" = 0.353 / 4.444
"mole fraction" = 0.079
