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Sodium chloride (NaCl) is commonly used to melt ice on roads during the winter. Calcium chloride (CaCl2) is sometimes used for this purpose too. Let us compare the effectiveness of equal masses of these two compounds in lowering the freezing point of water by calculating the freezing point depression of solutions containing 220. g of each salt in 1.00 kg of water. (An advantage of is that it acts more quickly because it is hygroscopic, that is, it absorbs moisture from the air to create a solution and begin the process. A disadvantage is that this compound is more costly.) Assume full dissociation of ionic compounds. Kfp(H2O)= -1.86 °C/m.

ΔTfp= _________°C for NaCl
ΔTfp=_________ °C for CaCl2

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Answer:

[tex]\Delta Tfp_{NaCl}= -14.0\°C\\\\\Delta Tfp_{CaCl_2}=-11.1\°C[/tex]

Explanation:

Hello!

In this case, since the freezing point depression caused by the addition of a solute, we use the following formula:

[tex]\Delta Tfp= i*m*Kfp[/tex]

Thus, we first need to compute the molality of each solute, as shown below:

[tex]m_{NaCl}=\frac{220.g/(58.44g/mol)}{1.00kg} =3.76m\\\\m_{CaCl_2}=\frac{220.g/(110.98g/mol)}{1.00kg} =1.98m[/tex]

Next, since NaCl has two ionic species, one Na⁺ and one Cl⁻, and CaCl₂ three, one Ca²⁺ and two Cl⁻, the van't Hoff's factors are 2 and 3 respectively, therefore the freezing point depressions turn out:

[tex]\Delta Tfp_{NaCl}= 2*3.76m*-1.86\°C/m=-14.0\°C\\\\\Delta Tfp_{CaCl_2}= 3*1.98m*-1.86\°C/m=-11.1\°C[/tex]

It means that CaCl₂ is still better than NaCl because produces involves a higher melting point for the ice, so it would melt faster.

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