Answer:
a) Cm= 3.9 m ; ΔTf= 14.51 ºC
b) Cm= 0.21 m ; ΔTf= 0.79ºC
Explanation:
In order to solve the problems, we have to remember that the molality (m) of a solution is equal to moles of solute in 1 kg of solvent.
m= mol solute/kg solvent
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
