Ch.4 - Chemical Quantities & Aqueous ReactionsWorksheetSee all chapters
All Chapters
Ch.1 - Intro to General Chemistry
Ch.2 - Atoms & Elements
Ch.3 - Chemical Reactions
BONUS: Lab Techniques and Procedures
BONUS: Mathematical Operations and Functions
Ch.4 - Chemical Quantities & Aqueous Reactions
Ch.5 - Gases
Ch.6 - Thermochemistry
Ch.7 - Quantum Mechanics
Ch.8 - Periodic Properties of the Elements
Ch.9 - Bonding & Molecular Structure
Ch.10 - Molecular Shapes & Valence Bond Theory
Ch.11 - Liquids, Solids & Intermolecular Forces
Ch.12 - Solutions
Ch.13 - Chemical Kinetics
Ch.14 - Chemical Equilibrium
Ch.15 - Acid and Base Equilibrium
Ch.16 - Aqueous Equilibrium
Ch. 17 - Chemical Thermodynamics
Ch.18 - Electrochemistry
Ch.19 - Nuclear Chemistry
Ch.20 - Organic Chemistry
Ch.22 - Chemistry of the Nonmetals
Ch.23 - Transition Metals and Coordination Compounds

Solution: What is the concentration of nitrate ions in a solution made by dissolving 14.3 g of Fe(NO3)3 (molar mass = 241.88 g mol-1) in enough water to form 0.135 L of solution?A. 2.30 M B. 0.766 MC. 0.435 MD.

Problem

What is the concentration of nitrate ions in a solution made by dissolving 14.3 g of Fe(NO3)3 (molar mass = 241.88 g mol-1) in enough water to form 0.135 L of solution?

A. 2.30 M 

B. 0.766 M

C. 0.435 M

D. 1.31 M

E. 4.35 x 10-4 M


Solution

We’re being asked to calculate the molarity (M) of nitrate ions in a Fe(NO3)3 solution.


Recall that molarity is the ratio of the moles of solute and the volume of solution (in liters). In other words:


Molarity (M) = moles of soluteLiters of solution


We first need to determine the number of moles of Fe(NO3)3.


We’re given the mass and molar mass of Fe(NO3)3 so we calculate using the equation below. 


Molar Mass = massmol mol= massMolar Mass mol= 14.3  g  241.88 g/mol

mol = 0.059 mol Fe(NO3)3 


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