PART A A solution was prepared by dissolving 135.0 g of KCl in 225 g of water. C

ID: 573003 • Letter: P

Question

PART A

A solution was prepared by dissolving 135.0 g of KCl in 225 g of water.

Calculate the mole fraction of KCl. (The formula weight of KCl is 74.6 g/mol. The formula weight of water is 18.0 g/mol.)

Express the mole fraction of KCl to two decimal places.

PART B

A solution was prepared by dissolving 135.0 g of KCl in 225 g of water.

Calculate the molality of KCl. (The formula weight of KCl is 74.6 g/mol. The formula weight of water is 18.0 g/mol.)

Express the concentration of KCl in molality to two decimal places.

part-A

no of moles of KCl (n KCl) = W/G.F.Wt

= 135/74.6 = 1.81moles

no of moles of H2O (nH2O) = W/G.F.Wt

= 225/18 = 12.5 moles

mole fraction of KCl ( XKCl) = nKCl/nKCl + nH2O

= 1.81/1.81+12.5

= 1.81/14.31 = 0.13 >>>>answer

part-B

molality = W*1000/G.F.Wt * weight of solvent ( H2O) in g

= 135*1000/74.6*225 = 8.04 m

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