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Chapter 13: Problem 53

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

Short Answer

Expert verified
(a) Weigh out 1.34 g of solid KBr and dissolve it in approximately 0.7 L of water. Stir well and then add enough water to obtain a final volume of 0.75 L. (b) Weigh out 125 g of water and 2.68 g of solid KBr. Mix them together until the KBr is fully dissolved. (c) Weigh out 244.2 g of solid KBr and dissolve it in approximately 1.8 L of water. Stir well and then add enough water to obtain a final volume of 1.85 L. (d) Weigh out 5.74 g of solid KBr and dissolve it in approximately 0.3 L of water. Stir well, and then add enough water to obtain a final volume of 0.3217 L. Mix this solution with the AgNO3 solution to cause the precipitation of AgBr.

Step by step solution

01

(a) Calculating mass of KBr

To prepare a \(0.75 \mathrm{~L}\) of a \(1.5 \times 10^{-2} M \mathrm{KBr}\) solution, first calculate the moles of \(\mathrm{KBr}\) required using the formula: moles = molarity × volume. Moles of \(\mathrm{KBr} = 1.5 \times 10^{-2} \mathrm{M} \times 0.75 \mathrm{L} = 0.01125 \mathrm{mol}.\) Now, convert moles to mass, knowing the molar mass of \(\mathrm{KBr}\) is approximately 119 g/mol. Mass of \(\mathrm{KBr} = 0.01125 \mathrm{mol} \times 119 \mathrm{g/mol} = 1.34 \mathrm{g}.\)
02

(a) Preparing the solution

Weigh out 1.34 g of solid \(\mathrm{KBr}\) and dissolve it in approximately 0.7 L of water. Stir well and then add enough water to obtain a final volume of 0.75 L.
03

(b) Calculating mass of KBr

To prepare 125 g of a 0.180 m solution of \(\mathrm{KBr}\), first, calculate the mass of KBr to add to the water using the formula: mass of solute = molality × mass of solvent × molar mass of solute. Mass of \(\mathrm{KBr} = 0.180 \mathrm{m} \times 0.125 \mathrm{kg} \times 119 \mathrm{g/mol} = 2.68 \mathrm{g}.\)
04

(b) Preparing the solution

Weigh out 125 g of water and 2.68 g of solid \(\mathrm{KBr}\). Mix them together until the \(\mathrm{KBr}\) is fully dissolved.
05

(c) Calculating mass of KBr

Knowing the given volume (1.85 L) and density (1.10 g/mL) of the 12.0% by mass \(\mathrm{KBr}\) solution, first, calculate the total mass of the solution. Total mass of solution = 1.85 \mathrm{L} \times 1.10 \mathrm{g/mL} \times 1000 \mathrm{mL/L} = 2035 \mathrm{g}.$ Now, find the mass of \(\mathrm{KBr}\) in the solution. Mass of \(\mathrm{KBr} = 0.120 \times 2035 \mathrm{g} = 244.2 \mathrm{g}.\)
06

(c) Preparing the solution

Weigh out 244.2 g of solid \(\mathrm{KBr}\) and dissolve it in approximately 1.8 L of water. Stir well and then add enough water to obtain a final volume of 1.85 L.
07

(d) Moles of AgBr to be precipitated

Calculate the moles of AgBr that need to be precipitated: Moles of AgBr = \(16.0 \mathrm{~g}\times \frac{1 \mathrm{~mol}}{331.32 \mathrm{~g}} = 0.04826 \mathrm{~mol}\).
08

(d) Moles of KBr needed

For each mole of AgBr precipitated, one mole of \(\mathrm{KBr}\) is needed. Thus, we need 0.04826 mol of \(\mathrm{KBr}\).
09

(d) Calculating mass and volume of KBr

Calculate the mass of \(\mathrm{KBr}\) needed. Mass of \(\mathrm{KBr} = 0.04826 \mathrm{mol} \times 119 \mathrm{g/mol} = 5.74 \mathrm{g}.\) Calculate the volume of the \(0.150 \mathrm{M}\) solution needed. Volume of solution = \(\frac{0.04826 \mathrm{mol}}{0.150 \mathrm{M}}= 0.3217 \mathrm{L}\).
10

(d) Preparing the solution

Weigh out 5.74 g of solid \(\mathrm{KBr}\) and dissolve it in approximately 0.3 L of water. Stir well, and then add enough water to obtain a final volume of 0.3217 L. Mix this solution with the \(\mathrm{AgNO}_{3}\) solution to cause the precipitation of AgBr.

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