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Chapter 13: Problem 49

The density of acetonitrile \(\left(\mathrm{CH}_{3} \mathrm{CN}\right)\) is \(0.786 \mathrm{~g} / \mathrm{mL}\) and the density of methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) is \(0.791 \mathrm{~g} / \mathrm{mL}\). A solution is made by dissolving \(22.5 \mathrm{~mL} \mathrm{CH}_{3} \mathrm{OH}\) in \(98.7 \mathrm{~mL}\) CH \(_{3}\) CN. (a) What is the mole fraction of methanol in the solution? (b) What is the molality of the solution? (c) Assuming that the volumes are additive, what is the molarity of \(\mathrm{CH}_{3} \mathrm{OH}\) in the solution?

Short Answer

Expert verified
(a) The mole fraction of methanol in the solution is \(0.227\). (b) The molality of the methanol in the solution is \(7.17\, \text{mol/kg}\). (c) The molarity of methanol in the solution is \(4.59\,\text{M}\).

Step by step solution

01

Calculate Mass of Methanol and Acetonitrile

We need to find the mass of each substance before we can find their moles. We can use the density formula, which is: Density = mass/volume We can rearrange this formula to find the mass as follows: mass = density x volume For methanol (CH3OH): Mass = (0.791 g/mL) * (22.5 mL) = \(17.8\,\text{g}\) For acetonitrile (CH3CN): Mass = (0.786 g/mL) * (98.7 mL) = \(77.5\,\text{g}\)
02

Convert Mass to Moles

Next, we need to convert the mass of each substance to moles by using their molar masses. For methanol (CH3OH): molar mass = \(32.04\, \text{g/mol}\) Moles of methanol = \( \frac{17.8 \,\text{g}}{32.04 \, \text{g/mol}} = 0.556\,\text{mol} \) For acetonitrile (CH3CN): molar mass = \(41.05\, \text{g/mol}\) Moles of acetonitrile = \( \frac{77.5\,\text{g}}{41.05\, \text{g/mol}} = 1.89\,\text{mol} \)
03

Calculate Mole Fraction and Molality

To find mole fraction of methanol (X_CH3OH), we need to divide moles of methanol by the total moles of the solution. \[X_{CH3OH} = \frac{n\,\text{moles of CH3OH}}{n\,\text{moles of CH3OH} + n\,\text{moles of CH3CN}} = \frac{0.556}{0.556+1.89} = \frac{0.556}{2.446} = 0.227\] Molality (m) is defined as moles of solute (in this case, methanol) divided by mass of solvent (in this case, acetonitrile) in kilograms. \[m = \frac{n\,\text{moles of CH3OH}}{Mass\,of\,CH3CN\,(kg)} = \frac{0.556}{0.0775} = 7.17\, \text{mol/kg}\] Answer (a): The mole fraction of methanol in the solution is \(0.227\). Answer (b): The molality of the methanol in the solution is \(7.17\, \text{mol/kg}\).
04

Calculate Molarity of Methanol

As given, we should assume that the volumes are additive. We can now find the total volume and convert it to liters: \(Total\,Volume = 22.5\,\text{mL}\,CH3OH + 98.7\,\text{mL}\,CH3CN = 121.2\,\text{mL} = 0.1212\,\text{L}\) Now, we can find the molarity (M) which is defined as moles of solute divided by the total volume of the solution in liters: \[ M = \frac{n\,\text{moles of CH3OH}}{Total\, Volume \,(\text{L})} = \frac{0.556}{0.1212} = 4.59\, \text{M}\] Answer (c): The molarity of methanol in the solution is \(4.59\,\text{M}\).

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Most popular questions from this chapter

(a) Many proteins that remain homogeneously distributed in water have molecular masses in the range of 30,000 amu and larger. In what sense is it appropriate to consider such suspensions to be colloids rather than solutions? Explain. (b) What general name is given to a colloidal dispersion of one liquid in another? What is an emulsifying agent?

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

List the following aqueous solutions in order of decreasing freezing point: \(0.040 \mathrm{~m}\) glycerin \(\left(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}_{3}\right), 0.020 \mathrm{~m} \mathrm{KBr}\), \(0.030 \mathrm{~m}\) phenol \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}\right)\)

Which of the following in each pair is likely to be more soluble in water: (a) cyclohexane \(\left(\mathrm{C}_{6} \mathrm{H}_{12}\right)\) or glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) (Figure 13.12 ); (b) propionic acid \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\right)\) or sodium propionate \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COONa}\right) ;\) (c) HCl or ethyl chloride \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}\right) ?\) Explain in each case.

How does increasing the concentration of a nonvolatile solute in water affect the following properties: (a) vapor pressure, (b) freezing point, (c) boiling point; (d) osmotic pressure?
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