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Chapter 13: Problem 74

Using data from Table \(13.3,\) calculate the freezing and boiling points of each of the following solutions: (a) \(0.25 \mathrm{~m}\) glucose in ethanol; (b) \(20.0 \mathrm{~g}\) of decane, \(\mathrm{C}_{10} \mathrm{H}_{22}\), in \(50.0 \mathrm{~g} \mathrm{CHCl}_{3} ;\) (c) \(3.50 \mathrm{~g}\) \(\mathrm{NaOH}\) in \(175 \mathrm{~g}\) of water, (d) 0.45 mol ethylene glycol and \(0.15 \mathrm{~mol} \mathrm{KBr}\) in \(150 \mathrm{~g} \mathrm{H}_{2} \mathrm{O}\)

Short Answer

Expert verified
3 for ethanol. \(ΔT_f = K_f * m = 1.99 * 0.25 = 0.498 K\) \(ΔT_b = K_b * m = 1.22 * 0.25 = 0.305 K\) New freezing point = Original freezing point - ΔT_f = -114.14 - 0.498 = -114.638°C New boiling point = Original boiling point + ΔT_b = 78.37 + 0.305 = 78.675°C The freezing and boiling points of the 0.25 m glucose in ethanol solution are -114.638°C and 78.675°C, respectively.

Step by step solution

01

(a) Solution Properties

First, we need to find the molality of the glucose solution (m): Given mass of glucose = 0.25 mol Mass of solvent (ethanol) = 100 g (we assume 1 molal solution) molality (m) = moles of solute/mass of solvent (in kg) = 0.25 / (100/1000) = 0.25 mol/kg
02

(a) Calculate the Freezing and Boiling Points

We need to find the change in the freezing and boiling points using the respective depression and elevation constants provided in Table 13._makeConstraints

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Most popular questions from this chapter

The density of toluene \(\left(\mathrm{C}_{7} \mathrm{H}_{8}\right)\) is \(0.867 \mathrm{~g} / \mathrm{mL},\) and the density of thiophene \(\left(\mathrm{C}_{4} \mathrm{H}_{4} \mathrm{~S}\right)\) is \(1.065 \mathrm{~g} / \mathrm{mL}\). A solution is made by dissolving \(8.10 \mathrm{~g}\) of thiophene in \(250.0 \mathrm{~mL}\) of toluene. (a) Calculate the mole fraction of thiophene in the solution. (b) Calculate the molality of thiophene in the solution. (c) Assuming that the volumes of the solute and solvent are additive, what is the molarity of thiophene in the solution?

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Describe how you would prepare each of the following aqueous solutions: (a) \(1.50 \mathrm{~L}\) of \(0.110 \mathrm{M}\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\) solution, starting with solid \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4} ;\) (b) \(225 \mathrm{~g}\) of a solution that is \(0.65 \mathrm{~m}\) in \(\mathrm{Na}_{2} \mathrm{CO}_{3},\) starting with the solid solute; (c) \(1.20 \mathrm{~L}\) of a solution that is \(15.0 \% \mathrm{~Pb}\left(\mathrm{NO}_{3}\right)_{2}\) by mass (the density of the solution is \(1.16 \mathrm{~g} / \mathrm{mL}\) ), starting with solid solute; (d) a \(0.50 M\) solution of HCl that would just neutralize \(5.5 \mathrm{~g}\) of \(\mathrm{Ba}(\mathrm{OH})_{2}\) starting with \(6.0 \mathrm{M} \mathrm{HCl}\).

Consider two solutions, one formed by adding \(10 \mathrm{~g}\) of glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) to \(1 \mathrm{~L}\) of water and the other formed by adding \(10 \mathrm{~g}\) of sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) to \(1 \mathrm{~L}\) of water. Are the vapor pressures over the two solutions the same? Why or why not?
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