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Chapter 13: Problem 33

Calculate the mole fraction of the solute in the following aqueous solutions:(a) \(0.112 \mathrm{M} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) \((d=1.006 \mathrm{g} / \mathrm{mL})\) (b) \(3.20 \%\) ethanol,by volume \((d=0.993 \mathrm{g} / \mathrm{mL}\) pure \(\left.\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}, d=0.789 \mathrm{g} / \mathrm{mL}\right)\).

Short Answer

Expert verified
The mole fraction of the solute is \(0.0020\) in (a) and \(0.010\) in (b)

Step by step solution

01

Determine the number of moles of solute in (a)

e first part of the problem provides a solution in which the concentration of the solute, glucose, is given as \(0.112 M\). The molarity (M) is defined as the number of moles of the solute divided by the liters of the solution. Thus, one liter solution contains \(0.112 moles\) of glucose.
02

Calculate the number of moles of solvent in (a)

To find the number of moles of the solvent, we should know the mass and the molar mass of the solvent. The mass of the solvent can be found by subtracting the mass of solute from the total mass of the solution. Since this is an aqueous solution, the solvent is water, with a density of \(1.00 g/mL\) and a molar mass of \(18.015 g/mol\). Thus, in one liter of solution (\(1000 mL\)), the mass of water is \(1000 g\). Hence, the number of moles of water is \(1000 g / 18.015 g/mol = 55.5 moles\).
03

Calculate the mole fraction in (a)

The mole fraction is the ratio of the number of moles of the solute to the total number of moles in the solution. Hence, the mole fraction of glucose in the solution is \(0.112 moles / (0.112 moles + 55.5 moles) = 0.0020\). So the mole fraction of the solute in solution (a) is \(0.0020\)
04

Determine the number of moles of solute in (b)

In the second part of the problem, the concentration of the solute, ethanol, is given as 3.20% \(v/v\). This means that 3.20 mL of ethanol is present in 100 mL of solution. The molar mass of ethanol is \(46.07 g/mol\). The density of pure ethanol is \(0.789 g/mL\). Hence, the mass of ethanol in 100 mL of solution is \(3.20 mL * 0.789 g/mL = 2.52 g\). So, the number of moles of ethanol in 100 mL solution is \(2.52 g / 46.07 g/mol = 0.055 moles\).
05

Calculate the number of moles of solvent in (b)

To find the number of moles of the solvent, we should know the mass and the molar mass of the solvent. Since this is also an aqueous solution, the solvent is water, with a density of \(1 g/mL\) and a molar mass of \(18.015 g/mol\). The total volume of solution is 100 mL and we have 3.20 mL of ethanol in it. Hence, the volume of water in solution is \(100 mL - 3.20 mL = 96.8 mL\). So, the mass of water is \(96.8 g\) and the number of moles of water is \(96.8 g / 18.015 g/mol = 5.37 moles\).
06

Calculate the mole fraction in (b)

The mole fraction is the ratio of the number of moles of the solute to the total number of moles in the solution. Hence, the mole fraction of ethanol in the solution is \(0.055 moles / (0.055 moles + 5.37 moles) = 0.010\). So the mole fraction of the solute in solution (b) is \(0.010\).

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Most popular questions from this chapter

Calculate the vapor pressure at \(20^{\circ} \mathrm{C}\) of a saturated solution of the nonvolatile solute, urea, \(\mathrm{CO}\left(\mathrm{NH}_{2}\right)_{2},\) in methanol, \(\mathrm{CH}_{3} \mathrm{OH} .\) The solubility is \(17 \mathrm{g}\) urea/100 \(\mathrm{mL}\) methanol. The density of methanol is \(0.792 \mathrm{g} / \mathrm{mL}\), and its vapor pressure at \(20^{\circ} \mathrm{C}\) is \(95.7 \mathrm{mmHg}\).

Under an \(\mathrm{O}_{2}(\mathrm{g})\) pressure of \(1.00 \mathrm{atm}, 28.31 \mathrm{mL}\) of \(\mathrm{O}_{2}(\mathrm{g})\) dissolves in \(1.00 \mathrm{L} \mathrm{H}_{2} \mathrm{O}\) at \(25^{\circ} \mathrm{C} .\) What will be the molarity of \(\mathrm{O}_{2}\) in the saturated solution at \(25^{\circ} \mathrm{C}\) when the \(\mathrm{O}_{2}\) pressure is 3.86 atm? (Assume that the solution volume remains at \(1.00 \mathrm{L}\).)

When the stems of cut flowers are held in concentrated \(\mathrm{NaCl}(\mathrm{aq}),\) the flowers wilt. In a similar solution a fresh cucumber shrivels up (becomes pickled). Explain the basis of these phenomena.

The aqueous solubility at \(20^{\circ} \mathrm{C}\) of \(\mathrm{Ar}\) at \(1.00 \mathrm{atm}\) is equivalent to \(33.7 \mathrm{mL} \mathrm{Ar}(\mathrm{g}),\) measured at STP, per liter of water. What is the molarity of Ar in water that is saturated with air at 1.00 atm and \(20^{\circ} \mathrm{C}\) ? Air contains \(0.934 \%\) Ar by volume. Assume that the volume of water does not change when it becomes saturated with air.

NaCl(aq) isotonic with blood is \(0.92 \%\) NaCl (mass/volume). For this solution, what is (a) \(\left[\mathrm{Na}^{+}\right]\) (b) the total molarity of ions; (c) the osmotic pressure at \(37^{\circ} \mathrm{C} ;\) (d) the approximate freezing point? (Assume that the solution has a density of 1.005 g/mL.)
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