A solution was prepared by dissolving 27.0 of KCl in 225 g of water?

It always helps us to know whether you tried to work the problem and where you ran into difficulty.

First, I will assume that when you typed "27.0 of KCl " that you meant "27.0 g of KCl".

a) Calculate the mole fraction of KCl in the solution.

First, calculate the moles of all substances present in the mixture.

KCl: (27.0 g of KCl)/(74.5513 g/mol) = 0.3622 mol of KCl

H2O: (225 g of water)/(18.0153 g/mol) =12.4894 mol of H2O

Second, calculate the mole fraction of KCl.

mole fraction KCl = (mol KCl)/(total moles) = (mol KCl)/(mol KCl + mol H2O)

mole fraction KCl = (0.3622 mol of KCl)/(0.3622 mol of KCl + 12.4894 mol of H2O) = 0.3622/12.8516

mole fraction KCl = 0.0282

b) Calculate the molarity of KCl in the solution if the total volume of the solution is 239 mL.

molarity = moles/volume in liters

molarity of KCl = (0.3622 mol of KCl)/((239 mL)*(1 L/1000 mL))

molarity of KCl = 1.515 M (1.52 M to three sig figs)

c)Calculate the molality o KCl in the solution

molality = moles of solute/kg of solvent

molality of KCl = (0.3622 mol of KCl)/((225 g)*(1 kg/1000 g))

molality of KCl = 1.6096 m (1.61 m to three sig figs)