Calculate the molality for each of the following solutions. Then, calculate the freezing-point depression ΔTF = iKFcm produced by each of the salts. (Assume the density of water is 1.00 g/mL and KF = 1.86°C/m.) (a) 3.2 M NaCl (Assume the density of the solution is 1.00 g/mL.) Cm = WebAssign will check your answer for the correct number of significant figures. 3.9 Correct: Your answer is correct. m ΔTF = WebAssign will check your answer for the correct number of significant figures.14.5 Incorrect: Your answer is incorrect. Check the number of significant figures.°C (b) 24 g of KCl in 1.5 L of water Cm = WebAssign will check your answer for the correct number of significant figures. 0.21 Correct: Your answer is correct. m ΔTF = WebAssign will check your answer for the correct number of significant figures. 0.78 Incorrect: Your answer is incorrect. °C

a) In this case we have molarity, which is moles of solute in 1 liter of solution. We have to know how many kg of solvent (water) we have in 1 L of solution.

3.2 M NaCl= 3.2 mol NaCl/ 1 L solution

1 L solution= 1000 ml solution x 1.00 g/ml= 1000 g

A solution is composed by solute (NaCl) + solvent, so:

1000 g solution = g NaCl + g solvent

g NaCl= 3.2 mol NaCl x 58.44 g/mol= 187 g NaCl

g solvent= 1000 g - 187 g NaCl= 813 g= 0.813 kg

Cm= 3.2 g NaCl/0.813 kg solvent= 3.9 m

NaCl is an electrolyte and it dissociates in water in two ions: Na⁺ anc Cl⁻, si the van't Hoff factor (i) is 2.

ΔTf= i x KF x Cm= 2 x 1.86ºC/m x 3.9 m= 14.51ºC

b) In this case we have 24 g of solute in 1.5 L of solvent. We have to convert the liters of solvent to kg, and to convert the mass of solute to mol by using the molecular weight of KCl (74.55 g/mol):

24 g KCl x 1 mol KCl/74.55 g= 0.32 mol

1.5 L solvent= 1500 g solvent x 1.00 g/ml= 1500 g = 1.5 kg

Cm= 0.32 g KCl/1.5 kg solvent= 0.21 m

KCl is an electrolyte and when it dissolves in water, it dissociates in 2 ions: K⁺ and Cl⁻. For this, van't Hoff factor (i) is equal to 2.

ΔTf= i x KF x Cm= 2 x 1.86ºC x 0.21 m= 0.79ºC