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Chapter 4: Problem 67

Calculate the molarity of each of the following solutions: (a) \(29.0 \mathrm{~g}\) of ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) in \(545 \mathrm{~mL}\) of solution, (b) \(15.4 \mathrm{~g}\) of sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) in \(74.0 \mathrm{~mL}\) of solution, (c) \(9.00 \mathrm{~g}\) of sodium chloride \((\mathrm{NaCl})\) in \(86.4 \mathrm{~mL}\) of solution.

Short Answer

Expert verified
The molarities of the three solutions are approximately 1.15 M for ethanol, 0.608 M for sucrose, and 0.178 M for Sodium chloride.

Step by step solution

01

Convert mass to moles (Ethanol)

Ethanol (C2H5OH) has a molecular weight of approximately 46.07 g/mol. Convert the mass of ethanol (29.0 g) to moles using the formula \n\(Moles = mass / molecular weight\). Substituting the values we get moles of ethanol to be \(0.629 moles\) of ethanol.
02

Compute the Molarity (Ethanol)

We can compute the molarity by using the formula: \(Molarity = moles of solute / liters of solution\). Convert the volume of the solution from mL to liters, and substituting the values (0.629 moles and 0.545 liters) into the formula, we obtain the molarity as \(1.15 M\).
03

Convert mass to moles (Sucrose)

Sucrose (C12H22O11) has a molecular weight of approximately 342.3 g/mol. Convert the mass of sucrose (15.4 g) to moles using the formula \(Moles = mass / molecular weight\). Substituting the values gives \(\approx 0.045 moles\) of sucrose.
04

Compute the Molarity (Sucrose)

Apply the formula: \(Molarity = moles of solute / liters of solution\). After converting the volume of the solution from mL to liters (0.074 liters), and substitute the values (0.045 moles and 0.074 liters) into the formula, we obtain the molarity as \(\approx 0.608 M\).
05

Convert mass to moles (Sodium chloride)

Sodium chloride (NaCl) has a molecular weight of approximately 58.44 g/mol. Convert the mass of NaCl (9.00 g) to moles using the formula \(Moles = mass / molecular weight\). Substituting the values gives \(\approx 0.154 moles\) of NaCl.
06

Compute the Molarity (Sodium chloride)

Applying the molarity formula: \(Molarity = moles of solute / liters of solution\). Convert the volume of the solution from mL to liters (0.0.864 liters), and plug the values (0.154 moles and 0.0.864 liters) into the formula, we obtain the molarity as \(\approx 0.178 M\).

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Most popular questions from this chapter

Nitric acid is a strong oxidizing agent. State which of the following species is least likely to be produced when nitric acid reacts with a strong reducing agent such as zinc metal, and explain why: \(\mathrm{N}_{2} \mathrm{O}\) \(\mathrm{NO}, \mathrm{NO}_{2}, \mathrm{~N}_{2} \mathrm{O}_{4}, \mathrm{~N}_{2} \mathrm{O}_{5}, \mathrm{NH}_{4}^{+}\).

Give a chemical explanation for each of the following: (a) When calcium metal is added to a sulfuric acid solution, hydrogen gas is generated. After a few minutes, the reaction slows down and eventually stops even though none of the reactants is used up. (b) In the activity series, aluminum is above hydrogen, yet the metal appears to be unreactive toward steam and hydrochloric acid. (c) Sodium and potassium lie above copper in the activity series. In your explanation, discuss why \(\mathrm{Cu}^{2+}\) ions in a \(\mathrm{CuSO}_{4}\) solution are not converted to metallic copper upon the addition of these metals. (d) A metal M reacts slowly with steam. There is no visible change when it is placed in a pale green iron(II) sulfate solution. Where should we place \(\mathrm{M}\) in the activity series? (e) Before aluminum metal was obtained by electrolysis, it was produced by reducing its chloride \(\left(\mathrm{AlCl}_{3}\right)\) with an active metal. What metals would you use to produce aluminum in that way?

Explain how you would prepare potassium iodide (KI) by means of (a) an acid- base reaction and (b) a reaction between an acid and a carbonate compound.

A \(1.00-\mathrm{g}\) sample of a metal \(\mathrm{X}\) (that is known to form \(\mathrm{X}^{2+}\) ions ) was added to \(0.100 \mathrm{~L}\) of \(0.500 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\). After all the metal had reacted, the remaining acid required \(0.0334 \mathrm{~L}\) of \(0.500 \mathrm{M} \mathrm{NaOH}\) solution for neutralization. Calculate the molar mass of the metal and identify the element.

Sodium carbonate \(\left(\mathrm{Na}_{2} \mathrm{CO}_{3}\right)\) is available in very pure form and can be used to standardize acid solutions. What is the molarity of a HCl solution if \(28.3 \mathrm{~mL}\) of the solution are required to react with \(0.256 \mathrm{~g}\) of \(\mathrm{Na}_{2} \mathrm{CO}_{3} ?\)
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