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Chapter 13: Problem 45

Calculate the molality of each of the following solutions: (a) \(10.0 \mathrm{~g}\) of benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) dissolved in \(50.0 \mathrm{~g}\) of carbon tetrachloride \(\left(\mathrm{CCl}_{4}\right),(\mathbf{b}) 5.00 \mathrm{~g}\) of \(\mathrm{NaCl}\) dissolved in \(0.100 \mathrm{~L}\) of water.

Short Answer

Expert verified
In conclusion, the molality of solution (a) containing \(10.0\,g\) of benzene dissolved in \(50.0\,g\) of carbon tetrachloride is \(2.56\,m\), and the molality of solution (b) containing \(5.00\,g\) of NaCl dissolved in \(0.100\,L\) of water is \(0.86\,m\).

Step by step solution

01

Calculate moles of benzene

The molecular formula of benzene is \(C_6H_6\), so the molar mass of benzene is \((6 × 12.01)+(6 × 1.01)=78.12 g/mol\). Given 10.0 g of benzene, we can calculate the number of moles by: Moles of benzene = \(\frac{10.0\,\text{g}}{78.12\,\text{g/mol}} = 0.128\, \text{moles}\)
02

Convert mass of carbon tetrachloride

We must convert the mass of carbon tetrachloride from grams to kilograms: 50.0 g = \(50.0\, \times 10^{-3}\) kg = 0.050 kg
03

Calculate molality for (a)

Now, we can calculate the molality of the solution (a) as follows: Molality = \(\frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.128\,\text{moles}}{0.050\,\text{kg}} = 2.56\,m\) For (b):
04

Calculate moles of NaCl

The molecular formula of NaCl; the molar mass of NaCl is \((1 × 22.99)+(1 × 35.45)=58.44\,g/mol\). Given 5.00 g of NaCl, we can calculate the number of moles by: Moles of NaCl = \(\frac{5.00\,\text{g}}{58.44\,\text{g/mol}} = 0.086\,\text{moles}\)
05

Convert mass of water

We know that the density of water is about \(1\,\text{g/mL}\), so the mass of 0.100 L of water is: 0.100 L = 100.0 mL = 100.0 g 100.0 g = \(100.0\, \times 10^{-3}\) kg = 0.100 kg
06

Calculate molality for (b)

Now, we can calculate the molality of the solution (b) as follows: Molality = \(\frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.086\,\text{moles}}{0.100\,\text{kg}} = 0.86\,m\) In conclusion, the molality of solution (a) is \(2.56\,m\()), and the molality of solution (b) is \(0.86\,m\().

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Most popular questions from this chapter

You make a solution of a nonvolatile solute with a liquid solvent. Indicate if each of the following statements is true or false. (a) The freezing point of the solution is unchanged by addition of the solvent. (b) The solid that forms as the solution freezes is nearly pure solute. (c) The freezing point of the solution is independent of the concentration of the solute. (d) The boiling point of the solution increases in proportion to the concentration of the solute. (e) At any temperature, the vapor pressure of the solvent over the solution is lower than what it would be for the pure solvent.

The vapor pressure of pure water at \(70^{\circ} \mathrm{C}\) is $31.2 \mathrm{kPa}\(. The vapor pressure of water over a solution at \)70^{\circ} \mathrm{C}$ containing equal numbers of moles of water and glycerol \(\left(\mathrm{C}_{3} \mathrm{H}_{5}(\mathrm{OH})_{3}\right.\), a nonvolatile solute) is \(13.3 \mathrm{kPa}\). Is the solution ideal according to Raoult's law?

Calculate the number of moles of solute present in each of the following solutions: (a) \(255 \mathrm{~mL}\) of \(1.50 \mathrm{M} \mathrm{HNO}_{3}(a q)\) (b) \(50.0 \mathrm{mg}\) of an aqueous solution that is $1.50 \mathrm{~m} \mathrm{NaCl}$, (c) \(75.0 \mathrm{~g}\) of an aqueous solution that is \(1.50 \%\) sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) by mass.

List the following aqueous solutions in order of decreasing freezing point: \(0.040 \mathrm{~m}\) glycerin $\left(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}_{3}\right), 0.020 \mathrm{~m} \mathrm{KBr}$,

Suppose you had a balloon made of some highly flexible semipermeable membrane. The balloon is filled completely with a \(0.2 \mathrm{M}\) solution of some solute and is submerged in a 0.1 \(M\) solution of the same solute: Initially, the volume of solution in the balloon is \(0.25 \mathrm{~L}\). Assuming the volume outside the semipermeable membrane is large, as the illustration shows, what would you expect for the solution volume inside the balloon once the system has come to equilibrium through osmosis? [Section 13.5]
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