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Chapter 4: Problem 36

How much (a) ethanol, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}(d=0.789 \mathrm{g} / \mathrm{mL}),\) in liters, must be dissolved in water to produce \(200.0 \mathrm{L}\) of 1.65 \(\mathrm{M} \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH} ?\) (b) concentrated hydrochloric acid solution \((36.0 \%\) HCl by mass; \(d=1.18 \mathrm{g} / \mathrm{mL}),\) in milliliters, is required to produce 12.0 L of 0.234 M HCl?

Short Answer

Expert verified
a) 19.3 liters of Ethanol are required. b) 241.3 milliliters of HCl are required.

Step by step solution

01

Calculation for part a)

First, calculate the number of moles of ethanol in 200.0 L solution of 1.65 M ethanol using the formula: moles = Molarity * Volume(L). Substitute the given values and get \(moles = 1.65 mol/L * 200.0 L = 330.0 mol\) of ethanol. Then, convert this to grams using the molecular weight of \(C_2H_5OH = 46.07 g/mol\), by multiplying the number of moles with the molar mass, \(mass = 330.0 mol * 46.07 g/mol = 15202.1 g\). This is the mass of ethanol required.
02

Convert the mass to volume for part a)

Then, convert the mass of ethanol to volume (in liters) using the density given, \(d = 0.789 g/mL\). So, \(volume = mass / density = 15202.1 g / 0.789 g/mL = 19274.9 mL = 19.3 L (to 2 decimal places)\). This is the volume of ethanol that must be dissolved in water.
03

Calculation for part b)

First, calculate the number of moles in 12.0 L solution of 0.234 M HCl using the formula: moles = molarity * volume(L). Substitute these values, and get \(moles = 0.234 mol/L * 12.0 L = 2.808 mol\) of HCl. Since the concentrated HCl solution is 36% by mass, it means 100g of the solution contains 36g of HCl. Its molar mass is 36.5 g/mol, and therefore, the moles of HCl in 100g of solution is \(36.0 g / 36.5 g/mol = 0.986 mol\). Therefore, 1g of the solution contains \(0.986 mol /100 = 0.00986 mol\).
04

Calculation of volume for part b)

To get the grams of the solution that will provide the required moles of HCl, divide the number of moles by moles per gram: \(grams = 2.808 mol / 0.00986 mol/g = 284.7 g\). Convert this mass to volume using the given density \(d = 1.18 g/mL\), so \(volume = mass/density = 284.7g/1.18 g/mL = 241.3 mL\). Hence, that's the volume required of the concentrated HCl solution.

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Most popular questions from this chapter

Baking soda, \(\mathrm{NaHCO}_{3}\), is made from soda ash, a common name for sodium carbonate. The soda ash is obtained in two ways. It can be manufactured in a process in which carbon dioxide, ammonia, sodium chloride, and water are the starting materials. Alternatively, it is mined as a mineral called trona (left photo). Whether the soda ash is mined or manufactured, it is dissolved in water and carbon dioxide is bubbled through the solution. Sodium bicarbonate precipitates from the solution. As a chemical analyst you are presented with two samples of sodium bicarbonate-one from the manufacturing process and the other derived from trona. You are asked to determine which is purer and are told that the impurity is sodium carbonate. You decide to treat the samples with just sufficient hydrochloric acid to convert all the sodium carbonate and bicarbonate to sodium chloride, carbon dioxide, and water. You then precipitate silver chloride in the reaction of sodium chloride with silver nitrate. A \(6.93 \mathrm{g}\) sample of baking soda derived from trona gave \(11.89 \mathrm{g}\) of silver chloride. A \(6.78 \mathrm{g}\) sample from manufactured sodium carbonate gave \(11.77 \mathrm{g}\) of silver chloride. Which sample is purer, that is, which has the greater mass percent \(\mathrm{NaHCO}_{3} ?\)

The incomplete combustion of gasoline produces \(\mathrm{CO}(\mathrm{g})\) as well as \(\mathrm{CO}_{2}(\mathrm{g}) .\) Write an equation for \((\mathrm{a})\) the complete combustion of the gasoline component octane, \(\mathrm{C}_{8} \mathrm{H}_{18}(\mathrm{l}),\) and \((\mathrm{b})\) incomplete combustion of octane with \(25 \%\) of the carbon appearing as \(\mathrm{CO}(\mathrm{g})\)

A \(25.0 \mathrm{mL}\) sample of \(\mathrm{HCl}(\mathrm{aq})\) is diluted to a volume of 500.0 mL. If the concentration of the diluted solution is found to be \(0.085 \mathrm{M} \mathrm{HCl}\), what was the concentration of the original solution?

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 \(\mathrm{Cl}, \mathrm{H},\) \(\mathrm{O}, \mathrm{Al}, \mathrm{N} .\)]

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
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