Various Types of Solutions

Slides:



Advertisements
Similar presentations
SOLUTIONS Chapter 15.
Advertisements

Chapter 11 Properties of Solutions AP*. AP Learning Objectives  LO 1.16 The student can design and/or interpret the results of an experiment regarding.
1.7 International System of Units (SI). 1.7 Volume – SI derived unit for volume is cubic meter (m 3 ) 1 cm 3 = (1 x m) 3 = 1 x m 3 1 dm 3.
Concentration. Concentration Particles per volume Can be in grams per litre but chemists usually express concentration in moles per litre This are related.
Solution Concentration. Calculations of Solution Concentration: Mass Percent Mass percent Mass percent is the ratio of mass units of solute to mass units.
Molarity and Molality.
Concentration.
Chapt. 12 Solutions Sec. 1 Concentration. Units of Concentration I solvent + solute.
Chpt 11 - Solutions Concentrations Energy of solutions Solubility Colligative Properties HW: Chpt 11 - pg. xxx-xxx, #s Due Mon Dec. 14.
M OLARITY Solutions. M OLARITY (M) Is the number of moles of solute dissolved per Liter of solution. Also know as molar concentration. Unit M is read.
1 Concentration of Solute The amount of solute in a solution is given by its concentration The amount of solute in a solution is given by its concentration.
Example A MgSO 4 (FW= g/mol) aqueous solution has a weight fraction of 0.2. What is the molality of the solution? Solution A 0.2.
Warm-Up! 1. What is the molarity of an aqueous solution containing 40 g of NaOH in 1.5 L of solution? 2. How many grams of NaCl should be dissolved in.
Properties of Solutions
Solutions Molality. Molality (m) A concentration that expresses the moles of solute in 1 kg of solvent Molality (m) = moles of solute 1 kg solvent.
Notes 15.2 Describing Solution Composition. Mass Percent Mass percent= mass of solute X 100 mass of solution = grams of solute X 100 grams of solute +
Concentration Units: Terms like “dilute” and “concentrated” are not specific. Percent by Mass: Mass % = mass of solute x 100 Total mass of solution Recall:
Chapter 20 Concentration. Molarity (M) Moles of solute per liter of solution. Molarity = moles of solute liters of solution.
Percent by mass, mole fraction, molarity, and molality
% by Mass Another way to measure the concentration of a solution % by mass = mass solute x 100 mass solution Solution = solute + solvent.
CONCENTRATION OF SOLUTIONS. Solute + The amount of solution can be expressed by: - mass m (g, kg) or - volume V (cm 3, mL, dm 3, L, m 3 ) m = V x  -
Chapter 20 Concentration. Molarity (M) Moles of solute per liter of solution. Molarity = moles of solute liters of solution.
 One liter of 5.0 M HCl contains how many moles of HCl? M = 5.0 mol= ? L = 1 L x = 5 mol.
Solution Concentration. Concentration Describes the amount of solute dissolved in a specific amount of solvent.
Chapter Units of Concentration unitless Definition same.
Ch. 13/14: Solutions Describing a Solution’s Composition.
Solutions, Problems, Solutions, Problems. Does it ever end?
Solution. True or false 1. Whipping cream is an example of a solution. False, it is a colloid 2. Solutions show the tyndall effect. False, colloids do.
Concentration  A measure of how much solute is dispersed throughout the solvent  Molarity (M), molality (m), and mole fraction ( χ ), mass percent.
Molarity moles of solute Liters of solution Unit for molarity mole mol LiterL = M.
Solutions Part II: Concentration Units. Percent solutions.
Molar Concentration. Molarity is the number of moles of solute that can dissolve in 1 L of solution. Molar concentration (mol/L) = Amount of solute (mol)
1 Concentration Day 2: Chapter 14. CONCENTRATION The amount of solute present in a set amount of solvent or total solution. The amount of solute present.
Solution problems. If we dissolve 87 grams of silver into 200 grams of gold, the resulting alloy is what percent silver?
Chapter 11 Properties of Solutions. Copyright © Cengage Learning. All rights reserved 2 Solution – a homogeneous mixture. Solute – substance being dissolved.
Chapter 11 Properties of Solutions. Section 11.1 Solution Composition Copyright © Cengage Learning. All rights reserved 2 Various Types of Solutions.
Solution Concentrations The measured amount of solute in a given amount of solvent or solution.
ACT Prep Passage V 5 minutes Start when the bell rings
Molality (m) A concentration that expresses the
Molarity Thornburg 2014.
Solution Concentration
Solution Concentration Units
Chapter 11 Properties of Solutions
Concentration of Solute
Molarity (M): State the ratio between the number of moles of solute & the volume of solution (in liters). Molarity (M) =
Solution Concentration Units
DO NOW Pick up Notes. Get out Solubility handout and Concept Review handout.
LO 1.16 The student can design and/or interpret the results of an experiment regarding the absorption of light to determine the concentration of an absorbing.
Various Types of Solutions
To Do… OWL: due Wednesday (2/21) and Friday (2/23).
Solutions Molarity ICS.
Clicker #1 You add mL of water to mL of a 2.00 M sugar solution. How many of the following will change? I. total volume of the solution II. moles.
Molarity & Dilution.
Ch. 3 & 7 – The Mole II. Concentration (p )
Solutions Section 3: Dilutions Dilutions Calculations print 1-3,5-7.
Concentration of Solute
Clicker #1 Which of the three solutions is the most concentrated?
Various Types of Solutions
Clicker #1 Which of the three solutions is the most concentrated?
Chapter 13.5 Expressing solution concentration
II. Molarity.
Clicker #1 Which of the three solutions is the most concentrated?
LO 1.16 The student can design and/or interpret the results of an experiment regarding the absorption of light to determine the concentration of an absorbing.
8.2 Measuring Concentration Objective 2
Molarity Calculate the concentration of a solute in terms of grams per liter, molarity, and percent composition.
Clicker #1 You add mL of water to mL of a 2.00 M sugar solution. How many of the following will change? I. total volume of the solution II. moles.
Concentration of Solutions :
Clicker #1 You add mL of water to mL of a 2.00 M sugar solution. How many of the following will change? I. total volume of the solution II. moles.
Molarity (M): State the ratio between the number of moles of solute & the volume of solution (in liters). Molarity (M) =
Table of Contents Chapter 15: Solutions.
Presentation transcript:

Various Types of Solutions Copyright © Cengage Learning. All rights reserved

Solution Composition Copyright © Cengage Learning. All rights reserved

Molarity Copyright © Cengage Learning. All rights reserved

EXERCISE! You have 1.00 mol of sugar in 125.0 mL of solution. Calculate the concentration in units of molarity. 1.00 mol / (125.0 / 1000) = 8.00 M Copyright © Cengage Learning. All rights reserved

EXERCISE! You have a 10.0 M sugar solution. What volume of this solution do you need to have 2.00 mol of sugar? 2.00 mol / 10.0 M = 0.200 L Copyright © Cengage Learning. All rights reserved

EXERCISE! Consider separate solutions of NaOH and KCl made by dissolving 100.0 g of each solute in 250.0 mL of solution. Calculate the concentration of each solution in units of molarity. [100.0 g NaOH / 39.998 g/mol] / [250.0 / 1000] = 10.0 M NaOH [100.0 g KCl / 74.55 g/mol] / [250.0 / 1000] = 5.37 M KCl Copyright © Cengage Learning. All rights reserved

Mass Percent Copyright © Cengage Learning. All rights reserved

EXERCISE! What is the percent-by-mass concentration of glucose in a solution made my dissolving 5.5 g of glucose in 78.2 g of water? [5.5 g / (5.5 g + 78.2 g)] × 100 = 6.6% Copyright © Cengage Learning. All rights reserved

Mole Fraction Copyright © Cengage Learning. All rights reserved

EXERCISE! A solution of phosphoric acid was made by dissolving 8.00 g of H3PO4 in 100.0 mL of water. Calculate the mole fraction of H3PO4. (Assume water has a density of 1.00 g/mL.) 8.00 g H3PO4 × (1 mol / 97.994 g H3PO4) = 0.0816 mol H3PO4 100.0 mL H2O × (1.00 g H2O / mL) × (1 mol / 18.016 g H2O) = 5.55 mol H2O Mole Fraction (H3PO4) = 0.0816 mol H3PO4 / [0.0816 mol H3PO4 + 5.55 mol H2O] = 0.0145 Copyright © Cengage Learning. All rights reserved

Molality Copyright © Cengage Learning. All rights reserved

EXERCISE! A solution of phosphoric acid was made by dissolving 8.00 g of H3PO4 in 100.0 mL of water. Calculate the molality of the solution. (Assume water has a density of 1.00 g/mL.) 8.00 g H3PO4 × (1 mol / 97.994 g H3PO4) = 0.0816 mol H3PO4 100.0 mL H2O × (1.00 g H2O / mL) × (1 kg / 1000 g) = 0.1000 kg H2O Molality = 0.0816 mol H3PO4 / 0.1000 kg H2O] = 0.816 m Copyright © Cengage Learning. All rights reserved