solutions: molarity
DESCRIPTION
Solutions: Molarity. Chemistry Ms. Piela. Solutions: Molarity. I. Concentration of Solutions. A. Concentration – measure of the amount of solute that is dissolved in a given amount of solvent. B. Dilute Solution – contains a lower concentration of solute. - PowerPoint PPT PresentationTRANSCRIPT
Solutions: MolarityChemistryMs. Piela
• A. Concentration – measure of the amount of solute that is dissolved in a given amount of solvent
• I. Concentration of Solutions•Solutions: Molarity
• B. Dilute Solution – contains a lower concentration of solute
• C. Concentrated Solution – contains a higher concentration of solute
• B. Sometimes referred to as a molar solution (i.e. 0.5 molar solution
• II Molarity• Solutions: Molarity
• A. The most important unit of concentration in chemistry
• The number of moles of solute dissolved per liter (L) of solution
• C. To determine molarity, DIVIDE the number of MOLES by the volume in LITERS
D. Example Problems
1. What is the molarity of a solution that contains 2.0 mol of glucose in 5.0 L of water?
0.4 M
• II Molarity• Solutions: Molarity
D. Example Problems2. A saline solution contains 0.90 g NaCl in exactly 0.100 L of solution. What is the molarity of the solution?
0.20L
• II Molarity• Solutions: Molarity
D. Example Problems
3. How many moles of solute are present in 1.5 L of 0.24 M Na2SO4?
0.36 mol
• II Molarity• Solutions: Molarity
•Solution Stoichiometry• Stoichiometry calculations can be done using molarity and volumes.• Since stoichiometry just depends on mole ratios, the goal is to find moles!
•Example Problem #1
•Calculate the number of mL of 2.00 M HNO3 solution required to react with 216 grams of Ag according to the equation.
3 Ag(s) + 4 HNO3 (aq) → 3 AgNO3 (aq) + NO(g) + 2 H2O(l)
•Example Problem #2
•Calculate in mL the volume of 0.500 M NaOH required to react with 3.0 grams of acetic acid (HC2H3O2). The equation is:
NaOH (aq) + HC2H3O2 (aq) → NaC2H3O2 (aq) + H2O(l)
•Example Problem #3
•Calculate the number of grams of AgCl formed when 0.200 L of 0.200 M AgNO3 reacts with an excess of CaCl2. The equation is:
2 AgNO3 (aq) + CaCl2 (aq) → 2 AgCl(s) + Ca(NO3)2 (aq)