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Chapter 13: Problem 98

Acetonitrile \(\left(\mathrm{CH}_{3} \mathrm{CN}\right)\) is a polar organic solvent that dissolves a wide range of solutes, including many salts. The density of a \(1.80 \mathrm{M}\) LiBr solution in acetonitrile is $0.826 \mathrm{~g} / \mathrm{cm}^{3}$. Calculate the concentration of the solution in (a) molality, (b) mole fraction of LiBr, (c) mass percentage of \(\mathrm{CH}_{3} \mathrm{CN}\).

Short Answer

Expert verified
The concentration of the LiBr solution in acetonitrile is: (a) 2.94 mol/kg in molality, (b) 0.1086 in mole fraction of LiBr, and (c) 73.56% in mass percentage of \(CH_{3}CN\).

Step by step solution

01

Calculate the molar mass of LiBr and Acetonitrile

First, we need to calculate the molar mass of LiBr and acetonitrile. LiBr: Li=6.939 g/mol, Br=79.904 g/mol, so LiBr = 6.939 + 79.904 = 86.843 g/mol Acetonitrile: C=12.01 g/mol, H=1.008 g/mol, N=14.007 g/mol, so CH3CN = 12.01 + 3*1.008 + 14.007 = 41.053 g/mol
02

Calculate the volume and mass of 1L solution

Now, we must determine the volume and mass of a 1-liter solution. Density = 0.826 g/cm³, so the mass of 1000 cm³ (1L) = density * volume = 0.826 g/cm³ * 1000 cm³ = 826 g
03

Calculate the moles of LiBr and Acetonitrile in 1L of solution

Next, we need to find the moles of LiBr and acetonitrile in 1L of solution. Molarity of LiBr = 1.80 mol/L Moles of LiBr in 1L solution = 1.80 mol Moles of Acetonitrile in 1L solution = (Mass of solution - Moles of LiBr × Molar mass of LiBr) / Molar mass of Acetonitrile Moles of Acetonitrile = (826 g - (1.80 mol × 86.843 g/mol)) / 41.053 g/mol ≈ 14.76 mol
04

Calculate the molality of the solution

Now we can calculate the molality of the solution (m), which is defined as moles of solute (LiBr) per kilogram of solvent (Acetonitrile). Molality = (moles of LiBr) / (mass of CH3CN in kg) Molality = (1.80 mol)/(14.76 mol * 41.053 g/mol * 0.001 kg/g) ≈ 2.94 mol/kg
05

Calculate the mole fraction of LiBr

Next, we calculate the mole fraction of LiBr (X_LiBr), which is the ratio of moles of LiBr to the total moles in the solution. Mole fraction of LiBr = (moles of LiBr) / (moles of LiBr + moles of Acetonitrile) X_LiBr = (1.80 mol)/(1.80 mol + 14.76 mol) ≈ 0.1086
06

Calculate the mass percentage of Acetonitrile

Finally, we need to calculate the mass percentage of acetonitrile in the solution, which is the ratio of the mass of acetonitrile to the total mass of the solution, expressed as a percentage. Mass percentage of Acetonitrile = (mass of Acetonitrile) / (mass of solution) * 100 Mass percentage of Acetonitrile = (14.76 mol * 41.053 g/mol) / (826 g) * 100 ≈ 73.56% The calculated values are: (a) Molality: 2.94 mol/kg (b) Mole fraction of LiBr: 0.1086 (c) Mass percentage of Acetonitrile: 73.56%

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Most popular questions from this chapter

Calculate the number of moles of solute present in each of the following aqueous solutions: (a) \(750 \mathrm{~mL}\) of \(0.120 \mathrm{M}\) \(\operatorname{SrBr}_{2},(\mathbf{b}) 70.0 \mathrm{~g}\) of $0.200 \mathrm{~m} \mathrm{KCl},(\mathbf{c}) 150.0 \mathrm{~g}\( of a solution that is \)5.75 \%$ glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) by mass.

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