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Chapter 4: Problem 34

What are the molarities of the following solutes? (a) aspartic acid \(\left(\mathrm{H}_{2} \mathrm{C}_{4} \mathrm{H}_{5} \mathrm{NO}_{4}\right)\) if \(0.405 \mathrm{g}\) is dissolved in enough water to make \(100.0 \mathrm{mL}\) of solution (b) acetone, \(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O},(d=0.790 \mathrm{g} / \mathrm{mL})\) if \(35.0 \mathrm{mL}\) is dissolved in enough water to make \(425 \mathrm{mL}\) of solution (c) diethyl ether, \(\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{O},\) if \(8.8 \mathrm{mg}\) is dissolved in enough water to make 3.00 L of solution

Short Answer

Expert verified
The molarities of the solutions are as follows: (a) Aspartic acid, 0.025 M; (b) Acetone, 1.02 M; (c) Diethyl ether, 0.00004 M

Step by step solution

01

Calculate the molarity of aspartic acid

While normally we would convert grams to moles using the molar mass of aspartic acid, we are given the total mass is 0.405g, so we can use that directly. Convert the volume of the solution to liters by dividing it by 1000. Therefore, the volume of the solution is 0.1L. Then use the definition of molarity, which is moles of solute per liter of solution. Given the density of aspartic acid is 1.66 g/cm^3, the molarity can be calculated as follows: \[ M = \frac{0.405 g}{(1.66 g/mol) \times 0.1 L} \]
02

Calculate the molarity of acetone

Calculate the number of moles of acetone by multiplying its volume by its density and dividing by its molar mass: \[ moles = \frac{35.0 mL \times (0.790 g/mL)}{(58.08 g/mol)} \] The volume of the solution is 425 mL, or 0.425 L. Therefore, the molarity of the acetone solution can be calculated as: \[ M = \frac{moles}{0.425 L} \]
03

Calculate the molarity of diethyl ether

Calculate the number of moles of diethyl ether by converting the mass of the solute from mg to g, and then dividing by the molar mass: \[ moles = \frac{8.8 mg \times (1 g/1000 mg)} {(74.12 g/mol)} \] The volume of the solution is 3.00 L. Therefore, the molarity of the diethyl ether solution can be calculated as: \[ M = \frac{moles}{3.00 L} \] The molarity for each solution has been determined.

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Most popular questions from this chapter

An essentially \(100 \%\) yield is necessary for a chemical reaction used to analyze a compound, but it is almost never expected for a reaction that is used to synthesize a compound. Explain this difference.

A method of lowering the concentration of \(\mathrm{HCl}(\mathrm{aq})\) is to allow the solution to react with a small quantity of Mg. How many milligrams of Mg must be added to \(250.0 \mathrm{mL}\) of \(1.023 \mathrm{M} \mathrm{HCl}\) to reduce the solution concentration to exactly \(1.000 \mathrm{M} \mathrm{HCl} ?\) $$ \mathrm{Mg}(\mathrm{s})+2 \mathrm{HCl}(\mathrm{aq}) \longrightarrow \mathrm{MgCl}_{2}(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{g}) $$

To prepare a solution that is \(0.50 \mathrm{M} \mathrm{KCl}\) starting with \(100.0 \mathrm{mL}\) of \(0.40 \mathrm{M} \mathrm{KCl},\) you should \((\mathrm{a})\) add \(20.0 \mathrm{mL}\) of water; (b) add 0.075 g KCl; (c) add 0.10 mol KCl; (d) evaporate \(20.0 \mathrm{mL}\) of water.

Solid calcium carbonate, \(\mathrm{CaCO}_{3}(\mathrm{s}),\) reacts with \(\mathrm{HCl}(\mathrm{aq})\) to form \(\mathrm{H}_{2} \mathrm{O}, \mathrm{CaCl}_{2}(\mathrm{aq}),\) and \(\mathrm{CO}_{2}(\mathrm{g}) .\) If a \(45.0 \mathrm{g}\) sample of \(\mathrm{CaCO}_{3}(\mathrm{s})\) is added to \(1.25 \mathrm{L}\) of \(\mathrm{HCl}(\mathrm{aq})\) that is \(25.7 \% \mathrm{HCl}\) by mass \((d=1.13 \mathrm{g} / \mathrm{mL})\) what will be the molarity of \(\mathrm{HCl}\) in the solution after the reaction is completed? Assume that the solution volume remains constant.

Under appropriate conditions, copper sulfate, potassium chromate, and water react to form a product containing \(\mathrm{Cu}^{2+},\) \(\mathrm{CrO}_{4}{^2}{^-},\) and \(\mathrm{OH}^{-}\) ions. Analysis of the compound yields \(48.7 \% \mathrm{Cu}^{2+}, 35.6 \% \mathrm{CrO}_{4}{^2}{-},\) and \(15.7 \% \mathrm{OH}^{-}\). (a) Determine the empirical formula of the compound. (b) Write a plausible equation for the reaction.
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