In the winter of \(1994,\) record low temperatures were registered throughout the United States. For example, in Champaign, Illinois, a record low of \(-29^{\circ} \mathrm{F}\) was registered. At this temperature can salting icy roads with \(\mathrm{CaCl}_{2}\) be effective in melting the ice? a. Assume \(i=3.00\) for \(\mathrm{CaCl}_{2}\) b. Assume the average value of \(i\) from Exercise 87 (The solubility of \(\mathrm{CaCl}_{2}\) in cold water is \(74.5 \mathrm{g}\) per \(100.0 \mathrm{g}\) water.

Short Answer

Expert verified
In this exercise, we were asked to determine if salting icy roads with CaCl2 can be effective in melting the ice at a record low temperature of -29°C. We calculated the molality of the solution and the freezing point depression for both cases: assuming i=3.00 and assuming the average value of i from Exercise 87. For Case a, the freezing point depression is 37.5 °C, meaning that melting of the ice is effective at temperatures as low as -37.5 °C. Since the record low temperature is -29 °C, salting the icy roads with CaCl2 will be effective in melting the ice at this temperature. For Case b, without knowing the average value of i, we cannot draw a conclusion. However, the method would be the same - calculate the freezing point depression and compare it with the record low temperature.

Step by step solution

01

Calculate the molality of the solution

Since we know the solubility of CaCl2, we can calculate the molality (m) of the solution. Molality is defined as moles of solute per kilogram of solvent. First, let's find the moles of CaCl2 present in 74.5 g: Molar mass of CaCl2 = 40.08 (Ca) + 2 * 35.45 (Cl) = 110.98 g/mol Moles of CaCl2 = (74.5 g) / (110.98 g/mol) = 0.671 mol Now let's find the molality using the mass of water: Mass of water = 100 g = 0.1 kg Molality (m) = moles of solute / mass of solvent (in kg) m = 0.671 mol / 0.1 kg = 6.71 mol/kg
02

Calculate freezing point depression for Case a

For Case a, we assume that i = 3.00 for CaCl2. We need the value of the molal freezing-point-depression constant (Kf) for water, which is 1.86°C/mol/kg. Now, let's calculate the freezing point depression using the formula: ∆Tf = i * Kf * m ∆Tf = 3.00 * 1.86 °C/mol/kg * 6.71 mol/kg ∆Tf = 37.5 °C
03

Calculate freezing point depression for Case b

For Case b, we assume the average value of i from Exercise 87 (which is not specified here). For the sake of completeness, we will denote this average value as i_avg. The calculation will be as follows: ∆Tf = i_avg * Kf * m
04

Analyze the results and draw conclusions

Now, let's analyze the results obtained from both cases. - For Case a, the freezing point depression is 37.5 °C. Since the normal freezing point of water is 0°C, this means that the melting of the ice is effective at temperatures as low as -37.5 °C. Therefore, since the record low temperature is -29 °C, salting the icy roads with CaCl2 will be effective in melting the ice at this temperature. - For Case b, we cannot conclude without knowing the average value of i. However, the method would be the same - calculate the freezing point depression and compare it with the record low temperature to determine the effectiveness of CaCl2 in melting the ice at -29 °C.

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Most popular questions from this chapter

An unknown compound contains only carbon, hydrogen, and oxygen. Combustion analysis of the compound gives mass percents of \(31.57 \%\) C and \(5.30 \%\) H. The molar mass is determined by measuring the freezing-point depression of an aqueous solution. A freezing point of \(-5.20^{\circ} \mathrm{C}\) is recorded for a solution made by dissolving \(10.56 \mathrm{g}\) of the compound in \(25.0 \mathrm{g}\) water. Determine the empirical formula, molar mass, and molecular formula of the compound. Assume that the compound is a nonelectrolyte.

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