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Solutions: Molarity

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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

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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

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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

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D. Example Problems 2. 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

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D. Example Problems 3. How many moles of solute are present in 1.5 L of 0.24 M Na 2 SO 4 ? 0.36 mol II Molarity Solutions: Molarity

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Solution Stoichiometry Stoichiometry calculations can be done using molarity and volumes. Since stoichiometry just depends on mole ratios, the goal is to find moles!

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Example Problem #1 Calculate the number of mL of 2.00 M HNO 3 solution required to react with 216 grams of Ag according to the equation. 3 Ag (s) + 4 HNO 3 (aq) → 3 AgNO 3 (aq) + NO (g) + 2 H 2 O (l)

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Example Problem #2 Calculate in mL the volume of 0.500 M NaOH required to react with 3.0 grams of acetic acid (HC 2 H 3 O 2 ). The equation is: NaOH (aq) + HC 2 H 3 O 2 (aq) → NaC 2 H 3 O 2 (aq) + H 2 O (l)

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Example Problem #3 Calculate the number of grams of AgCl formed when 0.200 L of 0.200 M AgNO 3 reacts with an excess of CaCl 2. The equation is: 2 AgNO 3 (aq) + CaCl 2 (aq) → 2 AgCl (s) + Ca(NO 3 ) 2 (aq)

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