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: A chemist prepares a solution of aluminum chloride (AICI 3) by measuring out 24.6 g of aluminum chloride into a 250. mL volumetric flask and filling the flask to the mark with water. Calculate the con

Solution: A chemist prepares a solution of aluminum chloride (AICI 3) by measuring out 24.6 g of aluminum chloride into a 250. mL volumetric flask and filling the flask to the mark with water. Calculate the con

Problem

A chemist prepares a solution of aluminum chloride (AICI 3) by measuring out 24.6 g of aluminum chloride into a 250. mL volumetric flask and filling the flask to the mark with water. Calculate the concentration in mol/L of the chemist's aluminum chloride solution. Be sure your answer has the correct number of significant digits.


Solution

Molarity can be calculated as:

For this problem, we are given the mass of AlCl3, we can calculate the moles by using its molar mass which is:

MM of AlCl3:

Al - 26.98 (1) = 26.98

Cl - 35.45 (3) = 106.36

26.98 + 106.35 = 133.34 g/mol

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