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Chapter 4: Problem 31

What are the molarities of the following solutes when dissolved in water? (a) \(2.92 \mathrm{mol} \mathrm{CH}_{3} \mathrm{OH}\) in 7.16 L of solution (b) \(7.69 \mathrm{mmol} \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) in \(50.00 \mathrm{mL}\) of solution (c) \(25.2 \mathrm{g} \mathrm{CO}\left(\mathrm{NH}_{2}\right)_{2}\) in \(275 \mathrm{mL}\) of solution

Short Answer

Expert verified
The molarities are (a) \(0.408 \mathrm{M}\), (b) \(0.154 \mathrm{M}\), and (c) \(1.524 \mathrm{M}\).

Step by step solution

01

Analyze the Given Values and Necessary Conversions

Check the given amounts of solute and the volume of the solutions. For each case, one needs to make sure that the amount of solute is in moles and the volume of solution is in liters. For (b), the amount of solute is given in millimoles, which need to be converted to moles. For (b) and (c), the volume of the solutions are given in milliliters, which need to be converted to liters. Therefore, the conversions that are necessary are \(1 \mathrm{mmol} = 10^{-3} \mathrm{mol}\) and \(1 \mathrm{mL} = 10^{-3} \mathrm{L}\).
02

Calculate the Molarity (a)

Molarity (M) is calculated as moles of solute divided by liters of solution. For (a), \(2.92 \mathrm{mol} \mathrm{CH}_{3} \mathrm{OH}\) in 7.16 L of solution gives a molarity of \(\frac{2.92 \mathrm{mol}}{7.16 \mathrm{L}} = 0.408 \mathrm{M}\).
03

Calculate the Molarity (b)

First convert the given millimoles to moles and milliliters to liters. Then calculate the molarity. This gives \(7.69 \mathrm{mmol} = 7.69 \times 10^{-3} \mathrm{mol}\) and \(50.00 \mathrm{mL} = 50.00 \times 10^{-3} \mathrm{L}\). Thus, the molarity is \(\frac{7.69 \times 10^{-3} \mathrm{mol}}{50.00 \times 10^{-3} \mathrm{L}} = 0.154 \mathrm{M}\).
04

Calculate the Molarity (c)

First, one needs to convert grams of solute to moles. The molar mass of \(\mathrm{CO}\left(\mathrm{NH}_{2}\right)_{2}\) is 60.06 g/mol. Hence, 25.2 g of \(\mathrm{CO}\left(\mathrm{NH}_{2}\right)_{2}\) is \(\frac{25.2 \mathrm{g}}{60.06 \mathrm{g/mol}} = 0.419 \mathrm{mol}\). Convert milliliters to liters and calculate the molarity. This gives \(275 \mathrm{mL} = 275 \times 10^{-3} \mathrm{L}\) and \(\frac{0.419 \mathrm{mol}}{275 \times 10^{-3} \mathrm{L}} = 1.524 \mathrm{M}\).

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Most popular questions from this chapter

A seawater sample has a density of \(1.03 \mathrm{g} / \mathrm{mL}\) and \(2.8 \% \mathrm{NaCl}\)l by mass. A saturated solution of \(\mathrm{NaCl}\) in water is \(5.45 \mathrm{M} \mathrm{NaCl} .\) How many liters of water would have to be evaporated from \(1.00 \times 10^{6} \mathrm{L}\) of the seawater before \(\mathrm{NaCl}\) would begin to crystallize? (A saturated solution contains the maximum amount of dissolved solute possible.)

The rocket boosters of the space shuttle Discovery, launched on July \(26,2005,\) used a fuel mixture containing primarily solid ammonium perchlorate, \(\mathrm{NH}_{4} \mathrm{ClO}_{4}(\mathrm{s}),\) and aluminum metal. The unbalanced chemical equation for the reaction is given below. \(\mathrm{Al}(\mathrm{s})+\mathrm{NH}_{4} \mathrm{ClO}_{4}(\mathrm{s}) \longrightarrow\) $$ \mathrm{Al}_{2} \mathrm{O}_{3}(\mathrm{s})+\mathrm{AlCl}_{3}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\mathrm{N}_{2}(\mathrm{g}) $$ What is the minimum mass of \(\mathrm{NH}_{4} \mathrm{ClO}_{4}\) consumed, per kilogram of \(\mathrm{Al}\), by the reaction of \(\mathrm{NH}_{4} \mathrm{ClO}_{4}\) and Al?[Hint: Balance the elements in the order \(\mathrm{Cl}, \mathrm{H},\) \(\mathrm{O}, \mathrm{Al}, \mathrm{N} .\)]

A 0.3126 g sample of oxalic acid, \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4},\) requires 26.21 mL of a particular concentration of \(\mathrm{NaOH}(\mathrm{aq})\) to complete the following reaction. What is the molarity of the \(\mathrm{NaOH}(\mathrm{aq}) ?\) \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(\mathrm{s})+2 \mathrm{NaOH}(\mathrm{aq}) \longrightarrow\) $$ \mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) $$

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.

Consider the reaction below. \(\mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{s})+2 \mathrm{HCl}(\mathrm{aq}) \longrightarrow \mathrm{CaCl}_{2}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) (a) How many grams of \(\mathrm{Ca}(\mathrm{OH})_{2}\) are required to react completely with \(415 \mathrm{mL}\) of \(0.477 \mathrm{M} \mathrm{HCl} ?\) (b) How many kilograms of \(\mathrm{Ca}(\mathrm{OH})_{2}\) are required to react with 324 L of a HCl solution that is 24.28\% HCl by mass, and has a density of \(1.12 \mathrm{g} / \mathrm{mL} ?\)
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