Question

in ice-cream making, the ingredients are kept below 0.0°C in an ice-salt bath.
A) assuming that NaCl dissolves completely and forms an ideal solution, what mass of it is needed to lower the melting point of 5.1 kg of ice to -4.2°C
B)given the same assumptions, what mass of CaCl2 is needed?

Answer 1

If a solution is treated as an ideal solution, the extent of freezing-point depression depends only on the solute concentration that can be estimated by a simple linear relationship with the cryoscopic constant

ΔT_F = K_F · m · i,

where:

• ΔT_F, the freezing-point depression, is defined as T_{F (pure solvent)} − T_{F (solution)}.
• K_F, the cryoscopic constant, which is dependent on the properties of the solvent, not the solute. For water, K_F = -1.853 K·kg/mol.
• m is the molality (moles solute per kilogram of solvent)
• i is the van 't Hoff factor (number of ion particles per individual molecule of solute, e.g. i = 2 for NaCl, 3 for CaCl₂).

Substituing in this formula we have for NaCl

-4.2 = -1.853 x moles of NaCl/5.1 Kg x 2

4.2 = 3.706 x moles of NaCl/5.1 Kg

moles of NaCl/5.1 Kg = 4.2/3.706

moles of NaCl/5.1 Kg = 1.133

moles of NaCl = 1.133 x 5.1

moles of NaCl = 5.78      Molecular weight of NaCl = 58.4 g/mol

Mass of NaCl = 5.78 x 58.4 = 337.5 g of NaCl is required to lower the temperature of 5.1 Kg ice to -4.2 ^oC

b) In the case of CaCl₂

-4.2 = -1.853 x moles of CaCl₂/5.1 Kg x 3

4.2 = 5.56 x moles of CaCl₂/5.1 Kg

moles of CaCl₂/5.1 Kg = 4.2/5.56

moles of CaCl₂/5.1 Kg = 0.755

moles of CaCl₂= 0.755 x 5.1

moles of CaCl₂ = 3.85    Molecular weight of CaCl₂= 110.98 g/mol

Mass of CaCl₂= 3.85 x 110.98 = 427.6 g of CaCl₂ is required to lower the temperature of 5.1 Kg ice to -4.2 ^oC