Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 4: Problem 102

What volume of \(0.0175 \mathrm{M} \mathrm{CH}_{3} \mathrm{OH}\) must be added to \(50.0 \mathrm{mL}\) of \(0.0248 \mathrm{M} \mathrm{CH}_{3} \mathrm{OH}\) so that the resulting solution has a molarity of exactly \(0.0200 \mathrm{M}\) ? Assume that the volumes are additive.

Short Answer

Expert verified
To calculate \(V_2\), substitute the known values into the equation from Step 3: \(V_2 = ((0.0248 M * 50.0 mL) - (0.0200 M * 50.0 mL)) / (0.0200 M - 0.0175 M)\). After calculation, you find that \(V_2 \approx 136.36 mL\). It indicates that about 136.36 mL of \(0.0175 M CH_{3}OH\) must be added to the solution.

Step by step solution

01

Label known variables

Firstly, identify and label all the known variables in the problem. Here, \(M_1 = 0.0248 M\), \(V_1 = 50.0 mL\), \(M_2 = 0.0175 M\), and \(M_f = 0.0200 M\).
02

Calculate the final volume

Since the volumes are additive, the final volume of the solution will be the volume of the first solution plus the volume of the second solution. Here, the second volume \(V_2\) is not known. So, let's denote the final volume as \(V_f = V_1 + V_2\).
03

Solve for the missing variable

Use the dilution equation \(M_1 V_1 + M_2 V_2 = M_f V_f\) to solve for \(V_2\). Firstly, substitute \(V_f\) on the right-hand side with \(V_1 + V_2\) to get \(M_1 V_1 + M_2 V_2 = M_f (V_1 + V_2)\). Then, rearrange the equation to find \(V_2: V_2 = (M_1 V_1 - M_f V_1) / (M_f - M_2)\). Next, substitute the known variables into this equation to solve for \(V_2\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Briefly describe (a) balancing a chemical equation; (b) preparing a solution by dilution; (c) determining the limiting reactant in a reaction.

Write chemical equations to represent the following reactions. (a) Calcium phosphate is heated with silicon dioxide and carbon, producing calcium silicate \(\left(\mathrm{CaSiO}_{3}\right)\) phosphorus ( \(\mathrm{P}_{4}\) ), and carbon monoxide. The phosphorus and chlorine react to form phosphorus trichloride, and the phosphorus trichloride and water react to form phosphorous acid. (b) Copper metal reacts with gaseous oxygen, carbon dioxide, and water to form green basic copper carbonate, \(\mathrm{Cu}_{2}(\mathrm{OH})_{2} \mathrm{CO}_{3}\) (a reaction responsible for the formation of the green patina, or coating, often seen on outdoor bronze statues). (c) White phosphorus and oxygen gas react to form tetraphosphorus decoxide. The tetraphosphorus decoxide reacts with water to form an aqueous solution of phosphoric acid. (d) Calcium dihydrogen phosphate reacts with sodium hydrogen carbonate (bicarbonate), producing calcium phosphate, sodium hydrogen phosphate, carbon dioxide, and water (the principal reaction occurring when ordinary baking powder is added to cakes, bread, and biscuits).

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

Azobenzene, an intermediate in the manufacture of dyes, can be prepared from nitrobenzene by reaction with triethylene glycol in the presence of \(\mathrm{Zn}\) and KOH. In one reaction, 0.10 L of nitrobenzene \((d=1.20 \mathrm{g} / \mathrm{mL})\) and \(0.30 \mathrm{L}\) of triethylene glycol \((d=1.12 \mathrm{g} / \mathrm{mL})\) yields \(55 \mathrm{g}\) azobenzene. What are the (a) theoretical yield, (b) actual yield, and (c) percent yield of this reaction? $$ 2 \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NO}_{2}+4 \mathrm{C}_{6} \mathrm{H}_{14} \mathrm{O}_{4} \frac{\mathrm{Zn}}{\mathrm{KOH}} $$ nitrobenzene triethylene glycol $$ \left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{N}\right)_{2}+4 \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{4}+4 \mathrm{H}_{2} \mathrm{O} $$ azobenzene

In the reaction of \(277 \mathrm{g} \mathrm{CCl}_{4}\) with an excess of \(\mathrm{HF}\) \(187 \mathrm{g} \mathrm{CCl}_{2} \mathrm{F}_{2}\) is obtained. What are the (a) theoretical, (b) actual, and (c) percent yields of this reaction? $$\mathrm{CCl}_{4}+2 \mathrm{HF} \longrightarrow \mathrm{CCl}_{2} \mathrm{F}_{2}+2 \mathrm{HCl}$$
See all solutions

Recommended explanations on Chemistry Textbooks

The Earths Atmosphere

Read Explanation

Nuclear Chemistry

Read Explanation

Kinetics

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Physical Chemistry

Read Explanation

Organic Chemistry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free