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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 short answer to the given problem is as follows: (a) The molality of the LiBr solution is 2.688 mol/kg. (b) The mole fraction of LiBr in the solution is 0.09935. (c) The mass percentage of CH₃CN in the solution is 81.07%.

Step by step solution

01

Calculate the mass of 1 L of solution

To calculate the mass of 1 L solution, we use the density formula: Density = \( \frac{mass}{volume} \) Rearranging, mass = Density × volume Given, Density = 0.826 g/cm³. We need to convert this density to g/L by multiplying with \(1000^3\). So, the density = 0.826 g/cm³ × 1000³ = 826 g/L. Now, we will find the mass of 1L solution: mass = Density × volume = 826 g/L × 1 L = 826 g
02

Calculate the mass of LiBr in 1 L of solution

To find the mass of the solute, LiBr, in 1 L of solution, we will use the given concentration of the LiBr solution: Concentration (M) = \( \frac{moles\,of\, solute}{volume\, of\, solution\, in\, L} \) Rearranging, moles of solute = Concentration × volume Given, Concentration = 1.80 M, volume = 1 L Now, we will find the moles of LiBr: moles of LiBr = 1.80 M × 1 L = 1.80 moles Next, we will find the molar mass of LiBr: Molar mass of LiBr (Li = 6.939 g/mol, Br = 79.904 g/mol) = 86.843 g/mol Now, we will find the mass of LiBr: mass of LiBr = moles of LiBr × molar mass of LiBr = 1.80 moles × 86.843 g/mol = 156.32 g
03

Calculate the mass of CH3CN in 1 L of solution

To find the mass of the solvent, CH3CN, we will subtract the mass of solute, LiBr from the total mass of the solution: mass of CH3CN = mass of solution - mass of LiBr = 826 g - 156.32 g = 669.68 g
04

Calculate the molality of the solution

To find the molality of the solution, we use the formula: Molality = \( \frac{moles\, of\, solute}{mass\, of\, solvent\, in\, kg} \) First, convert the mass of solvent, CH3CN to kg: mass of CH3CN in kg = 669.68 g × 0.001 = 0.66968 kg Molality = \( \frac{1.80\, moles\, of\, LiBr}{0.66968\, kg\, CH3CN} \) = 2.688 mol/kg (a) The molality of the solution is 2.688 mol/kg.
05

Calculate the mole fraction of LiBr

To find the mole fraction of LiBr, we use the formula: Mole fraction of LiBr = \( \frac{moles\, of\, LiBr}{total\, moles\, of\, solute\, and\, solvent} \) First, find the moles of CH3CN in the solution: Molar mass of CH₃CN = (12.01 + (3 × 1.01) + 14.01 + 12.01 g/mol)= 41.05 g/mol Moles of CH₃CN = \( \frac{mass\, of\, CH3CN}{molar\, mass\, of\, CH3CN} \) = \( \frac{669.68\, g}{41.05\, g/mol} \) = 16.316 moles Total moles of solute and solvent = moles of LiBr + moles of CH₃CN = 1.80 moles + 16.316 moles = 18.116 moles Mole fraction (χ) of LiBr = \( \frac{1.80}{18.116} \) = 0.09935 (b) The mole fraction of LiBr is 0.09935.
06

Calculate the mass percentage of CH3CN

To find the mass percentage of the solvent, CH3CN, we will use the formula: Mass percentage of CH₃CN = \( \frac{mass\, of\, CH3CN}{total\, mass\, of\, solution} \) × 100 Mass percentage of CH₃CN = \( \frac{669.68\, g}{826\, g} \) × 100 = 81.07 % (c) The mass percentage of CH3CN in the solution is 81.07 %.

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Most popular questions from this chapter

Indicate whether each statement is true or false: (a) The higher the temperature, the more soluble most gases are in water. (b) The higher the temperature, the more soluble most ionic solids are in water. (c) As you cool a saturated solution from high temperature to low temperature, solids start to crystallize out of solution if you achieve a supersaturated solution. (d) If you take a saturated solution and raise its temperature, you can (usually) add more solute and make the solution even more concentrated.

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