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Chapter 13: Problem 45

Calculate the molality of each of the following solutions: (a) 8.66 g of benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) dissolved in 23.6 \(\mathrm{g}\) of carbon tetrachloride \(\left(\mathrm{CCl}_{4}\right),(\mathbf{b}) 4.80 \mathrm{g}\) of NaCl dissolved in 0.350 \(\mathrm{L}\) of water.

Short Answer

Expert verified
The molality of the first solution (benzene in carbon tetrachloride) is 4.70 mol/kg, and the molality of the second solution (NaCl in water) is 0.235 mol/kg.

Step by step solution

01

Calculate the moles of benzene

To find the moles of benzene, we need to divide the mass of benzene by its molar mass. The molar mass of benzene (C6H6) is: (6 × 12.01 + 6 × 1.01) g/mol = 78.12 g/mol. moles of benzene = \(\frac{8.66}{78.12}\) moles of benzene = 0.1109 moles
02

Convert the mass of carbon tetrachloride to kg

Since molality is expressed in terms of moles of solute per kg of solvent, we need to convert the given mass of carbon tetrachloride (23.6 g) to kg. mass of carbon tetrachloride = 23.6 g = 0.0236 kg
03

Calculate the molality of benzene in carbon tetrachloride

Now, we can calculate the molality of benzene in carbon tetrachloride using the formula: molality (m) = \(\frac{moles \, of \, benzene}{mass \, of \, carbon \, tetrachloride \, (kg)}\) molality (m) = \(\frac{0.1109}{0.0236}\) molality (m) = 4.70 mol/kg The molality of benzene in carbon tetrachloride is 4.70 mol/kg. **Solution (b):**
04

Calculate the moles of NaCl

To find the moles of NaCl, we need to divide the mass of NaCl by its molar mass. The molar mass of NaCl is (22.99 + 35.45) g/mol = 58.44 g/mol. moles of NaCl = \(\frac{4.80}{58.44}\) moles of NaCl = 0.0822 moles
05

Convert the volume of water to mass (in kg)

We are given the volume of water (0.350 L) and need to convert it to mass (in kg) because molality is expressed in terms of mass. Since the density of water is 1 g/mL (or 1 kg/L), we can directly convert the volume to mass. mass of water = 0.350 L × 1 kg/L = 0.350 kg
06

Calculate the molality of NaCl in water

Now, we can calculate the molality of NaCl in water using the formula: molality (m) = \(\frac{moles \, of \, NaCl}{mass \, of \, water \, (kg)}\) molality (m) = \(\frac{0.0822}{0.350}\) molality (m) = 0.235 mol/kg The molality of NaCl in water is 0.235 mol/kg.

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Most popular questions from this chapter

The solubility of \(\mathrm{Cr}\left(\mathrm{NO}_{3}\right)_{3} \cdot 9 \mathrm{H}_{2} \mathrm{O}\) in water is 208 \(\mathrm{g}\) per 100 \(\mathrm{g}\) of water at \(15^{\circ} \mathrm{C}\) . A solution of \(\mathrm{Cr}\left(\mathrm{NO}_{3}\right)_{3} \cdot 9 \mathrm{H}_{2} \mathrm{O}\) in water at \(35^{\circ} \mathrm{C}\) is formed by dissolving 324 \(\mathrm{g}\) in 100 \(\mathrm{g}\) of water. When this solution is slowly cooled to \(15^{\circ} \mathrm{C},\) no precipitate forms. (a) Is the solution that has cooled down to \(15^{\circ}\) Cunsaturated, saturated, or supersaturated? (b) You take a metal spatula and scratch the side of the glass vessel that contains this cooled solution, and crystals start to appear. What has just happened? (c) At equilibrium, what mass of crystals do you expect to form?

Describe how you would prepare each of the following aqueous solutions, starting with solid KBr: (a) 0.75 L of \(1.5 \times 10^{-2} M \mathrm{KBr},\) (b) 125 \(\mathrm{g}\) of \(0.180 \mathrm{m} \mathrm{KBr},(\mathbf{c}) 1.85 \mathrm{L}\) of a solution that is 12.0\(\% \mathrm{KBr}\) by mass (the density of the solution is 1.10 \(\mathrm{g} / \mathrm{mL}\) , ( \(\mathrm{d}\) ) a 0.150 \(\mathrm{M}\) solution of \(\mathrm{KBr}\) that contains just enough KBr to precipitate 16.0 \(\mathrm{g}\) of AgBr from a solution containing 0.480 mol of \(\mathrm{AgNO}_{3} .\)

Carbon disulfide \(\left(\mathrm{CS}_{2}\right)\) boils at \(46.30^{\circ} \mathrm{C}\) and has a density of 1.261 \(\mathrm{g} / \mathrm{mL}\) . (a) When 0.250 \(\mathrm{mol}\) of a nondissociating solute is dissolved in 400.0 \(\mathrm{mL}\) of \(\mathrm{CS}_{2},\) the solution boils at \(47.46^{\circ} \mathrm{C} .\) What is the molal boiling-point-elevation constant for \(\mathrm{CS}_{2}\) ? (b) When 5.39 \(\mathrm{g}\) of a nondissociating unknown is dissolved in 50.0 \(\mathrm{mL}\) of \(\mathrm{CS}_{2},\) the solution boils at \(47.08^{\circ} \mathrm{C} .\) What is the molar mass of the unknown?

(a) Does a 0.10 \(\mathrm{m}\) aqueous solution of NaCl have a higher bolling point, a lower boiling point, or the same boiling point as a 0.10 \(\mathrm{m}\) aqueous solution of \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} ?(\mathbf{b})\) The experimental boiling point of the NaCl solution is lower than that calculated assuming that NaCl is completely dissociated in solution. Why is this the case?

The solubility of MnSO_ \(\cdot \mathrm{H}_{2} \mathrm{O}\) in water at \(20^{\circ} \mathrm{C}\) is 70 \(\mathrm{g}\) per 100 \(\mathrm{mL}\) of water. (a) Is a 1.22 \(\mathrm{M}\) solution of \(\mathrm{MnSO}_{4} \cdot \mathrm{H}_{2} \mathrm{O}\) in water at \(20^{\circ} \mathrm{C}\) saturated, supersaturated, or unsaturated? (b) Given a solution of MnSO_ \(\cdot \mathrm{H}_{2} \mathrm{O}\) of unknown concentration, what experiment could you perform to determine whether the new solution is saturated, supersaturated, or unsaturated?
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