×
Get Full Access to Chemistry: A Molecular Approach - 3 Edition - Chapter 4 - Problem 78e
Get Full Access to Chemistry: A Molecular Approach - 3 Edition - Chapter 4 - Problem 78e

×

# Solved: Write a molecular equation for the precipitation

ISBN: 9780321809247 1

## Solution for problem 78E Chapter 4

Chemistry: A Molecular Approach | 3rd Edition

• Textbook Solutions
• 2901 Step-by-step solutions solved by professors and subject experts
• Get 24/7 help from StudySoup virtual teaching assistants

Chemistry: A Molecular Approach | 3rd Edition

4 5 1 415 Reviews
23
4
Problem 78E

Problem 78E

Write a molecular equation for the precipitation reaction that occurs (if any) when each pair of aqueous solutions is mixed. If no reaction occurs, write “NO REACTION.”

a. sodium chloride and lead(II) acetate

b. potassium sulfate and strontium iodide

c. cesium chloride and calcium sulfide

d. chromium(III) nitrate and sodium phosphate

Step-by-Step Solution:
Step 1 of 3

Solution: Here, we are going to determine the molarity of a solution of glycerol made by dissolving 50 mL of glycerol in enough water to make 250 mL of solution. Step1: Molarity is defined as the number of moles of the solute(Here glycerol is the solute) in 1 litre of the solution. Mathematically, Number of moles of solute Molarity(M) = -------------------------------------- ------(1) Volume of solution in litres Step2: In order to calculate molarity of glycerol solution, we have to first calculate the number of moles of glycerol dissolved in the solution. Number of moles can be calculated by the following mathematical formula: Mass of the solute Number of moles(n) = ------------------------ Molar mass of the solute. Step3: o Given, density of glycerol at 15 C = 1.2656 g/mL Volume of the glycerol used = 50.00 mL Now, density is defined as mass per unit volume, i.e., Density = mass / volume Therefore, mass = density X volume = 1.2656 x 50.00 [(g/mL) X mL] = 63.28 g Thus, mass of the glycerol used = 63.28 g Molar mass of glycerol(C H38 392.08 g/mol Therefore, number of moles of glycerol = 63.28 / 92.08 = 0.687 mol Volume of the solution in litres = 250 mL = 250/1000 L = 0.250 L Putting these values in equation (1), we get, Molarity = 0.687 mol / 0.250 L = 2.748 M Thus , the molarity of the glycerol solution is 2.748 M. ------------

Step 2 of 3

Step 3 of 3

#### Related chapters

Unlock Textbook Solution