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Chapter 12: Problem 140

A student dissolves \(36.9 \mathrm{~g}\) of calcium nitrate in \(500.0 \mathrm{~g}\) of water. How many grams of glucose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\), would he have to dissolve in \(500.0 \mathrm{~g}\) of water to make the glucose solution have the same boiling point as the calcium nitrate solution?

Short Answer

Expert verified
The student would need to dissolve approximately \(121.6 \mathrm{~g}\) of glucose in \(500.0 \mathrm{~g}\) of water to make the glucose solution have the same boiling point as the calcium nitrate solution.

Step by step solution

01

Calculate molality of calcium nitrate solution

First, we need to find the molality of the calcium nitrate solution. Molality can be calculated using the formula: m = (moles of solute) / (mass of solvent in kg) Molecular weight of calcium nitrate, Ca(NO3)2, is: \(1 \times 40.08 \text{(Ca)} + 2 \times (1 \times 14.01 \text{(N)} + 3 \times 16.00 \text{(O)}) = 164.1 \mathrm{~g/mol}\) Moles of calcium nitrate = (36.9 g) / (164.1 g/mol) = 0.225 mol And we know that the mass of water (solvent) is 500.0 g (which is 0.500 kg). Now, we can calculate the molality: m = (0.225 mol) / (0.500 kg) = 0.450 mol/kg
02

Calculate boiling point elevation of calcium nitrate solution

Now that we have the molality, we can find the boiling point elevation using the formula ΔT = Kb * m * i. Since calcium nitrate completely dissociates into 3 ions (Ca²⁺ and 2NO₃⁻), the van't Hoff factor i = 3. The boiling point elevation constant for water, Kb, is 0.52 °C/mol/kg. ΔT = Kb * m * i = (0.52 °C/mol/kg) * (0.450 mol/kg) * (3) = 0.702 °C
03

Calculate molality needed for glucose solution

To get the same boiling point elevation, we can find the molality needed for the glucose solution (since glucose doesn't dissociate, i = 1): m_glucose = ΔT / (Kb * i) = 0.702 °C / (0.52 °C/mol/kg * 1) = 1.35 mol/kg
04

Calculate mass of glucose needed

Now that we have the required molality for the glucose solution, we can find the mass of glucose needed. The molecular weight of glucose, C6H12O6, is: \(6 \times 12.01 \text{(C)} + 12 \times 1.01 \text{(H)} + 6 \times 16.00 \text{(O)} = 180.16 \mathrm{~g/mol}\) We know that the mass of water is still 500.0 g (or 0.500 kg). Therefore, the mass of glucose needed can be calculated using the molality: mass of glucose = molality * (mass of solvent in kg) * molecular weight = (1.35 mol/kg) * (0.500 kg) * (180.16 g/mol) = 121.6 g The student would need to dissolve approximately 121.6 grams of glucose in 500.0 grams of water to make the glucose solution have the same boiling point as the calcium nitrate solution.

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Most popular questions from this chapter

A student combines \(60.0 \mathrm{~mL}\) of \(0.250 \mathrm{M} \mathrm{NaOH}\) with \(60.0 \mathrm{~mL}\) of \(0.125 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\). What is the hydroxide ion molar concentration in the resulting solution?

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