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Get Full Access to Chemistry: A Molecular Approach - 3 Edition - Chapter 4 - Problem 75e
Get Full Access to Chemistry: A Molecular Approach - 3 Edition - Chapter 4 - Problem 75e

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# Complete and balance each equation. If no reaction occurs,

ISBN: 9780321809247 1

## Solution for problem 75E Chapter 4

Chemistry: A Molecular Approach | 3rd Edition

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Chemistry: A Molecular Approach | 3rd Edition

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Problem 75E

Complete and balance each equation. If no reaction occurs, write “NO REACTION.”

a. LiI(aq) + BaS(aq) →

b. KCl(aq) + CaS(aq) →

c. CrBr2(aq) + Na2CO3(aq) →

d. NaOH(aq) + FeCl3(aq) →

Step-by-Step Solution:
Step 1 of 3

Solution: Here, we are going to describe a process to prepare 250 mL of a 0.250 M sucrose solution. Also, we are going to describe a process to prepare 350.0 ml of 0.100 M sucrose solution starting with 3.00 L of 1.50 M sucrose solution. Step1: Molarity is defined as the number of moles of the solute(Here sucrose is the solute) in 1 litre of the solution. Mathematically, Number of moles of solute Molarity(M) = -------------------------------------- ------(1) Volume of solution in litres Step2: a) Given, volume of the sucrose solution = 250 mL = 250 / 1000 L = 0.250 L Molarity of the sucrose solution = 0.250 M Now, using the molarity expression, we get, Number of moles of sucrose(C H O ) = v12m22f 11ution in litres X Molarity C12H22O11 = 0.250 X 0.250 M = 0.0625 mol Now, 1 mole of C H O 12o22 m11 of C H O 12 22 11 Therefore, 0.0625 moles of C H O = 0.0625 X molar mass of C H O 12 22 11 12 22 11 = 0.0625 X 342.3 g [molar mass of C12 O22 11.3 g/mol] = 21.39 g So, to prepare 250 ml of a 0.250 M sucrose solution, we have to dissolve 21.39 g of solid sucrose in a little amount of water in a 250 mL volumetric flask and then make the water level upto the mark in the neck of the flask. Step3: In order to dilute a stock solution, we have to use the following dilution equation, M 1 1M V 2 2 -----------(2) Here, M an1V are t1 molarity and volume of the concentrated stock solution whereas M and 2 V are the molarity and volume of the diluted solution. 2 b) Again, for the second problem, let us consider the 3.00 L of 1.50 M solution of sucrose to be concentrated solution and the solution of sucrose to be prepared as dilute solution. Therefore, M (concentrated) = 1.50 M 1 V1concentrated) = M 2ilute) = 0.100 V (dilute) = 350 mL 2 Putting these values in the equation(2), we get, 1.50 M X V = 0.10 M X 350 mL 1.50 M X V = 31(M x mL) V1 35 / 1.50 [(M x mL)/ M] V1 23.33 mL So, in a clean 50 mL burette 1.50 M sucrose solution is filled. Then, 23.33 mL of this solution is dispensed in a 350.0 mL of volumetric flask. Then, water is added to the mark and mixed thoroughly. ------------------

Step 2 of 3

Step 3 of 3

##### ISBN: 9780321809247

This textbook survival guide was created for the textbook: Chemistry: A Molecular Approach, edition: 3. Since the solution to 75E from 4 chapter was answered, more than 482 students have viewed the full step-by-step answer. This full solution covers the following key subjects: sucrose, describe, PREPARE, starting, solid. This expansive textbook survival guide covers 82 chapters, and 9454 solutions. The answer to “?Complete and balance each equation. If no reaction occurs, write “NO REACTION.” a. LiI(aq) + BaS(aq) ?b. KCl(aq) + CaS(aq) ? c. CrBr2(aq) + Na2CO3(aq) ?d. NaOH(aq) + FeCl3(aq) ?” is broken down into a number of easy to follow steps, and 30 words. The full step-by-step solution to problem: 75E from chapter: 4 was answered by , our top Chemistry solution expert on 02/22/17, 04:35PM. Chemistry: A Molecular Approach was written by and is associated to the ISBN: 9780321809247.

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Complete and balance each equation. If no reaction occurs,