Step 1: Calculate the moles of sucrose.

Slides:



Advertisements
Similar presentations
CALCULATIONS INVOLVING COLLIGATIVE PROPERTIES
Advertisements

Using Freezing Point to determine Molar Mass Objectives: To measure freezing points of known concentrations of solutions To use freezing point to determine.
Calculations Involving Colligative Properties. Introduction We now understand colligative properties. To use this knowledge, we need to be able to predict.
Molarity and Molality.
Colligative Properties. Properties that depend upon the concentration of solute particles are called colligative properties. Generally these properties.
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.
1 Colligative Properties of Solutions Colligative properties are properties that depend only on the number of solute particles in solution and not on the.
Molality Section 15.2.
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.
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
Chapter 20 Concentration. Molarity (M) Moles of solute per liter of solution. Molarity = moles of solute liters of solution.
Solutions Homogeneous Mixture of 2 or more Substances.
Chapter 13 Section 5: Colligative Properties
Solution Concentration
Mini-Unit: Colligative Properties
COLLIGATIVE PROPERTIES OF SOLUTIONS. Colligative Properties  A property that depends only upon the number of solute particles, and not upon their identity.
Solutions Review Problems have solutions, solutions have problems.
Calculations Involving Colligative Properties. Objectives When you complete this presentation, you will be able to o calculate the molality of a solution.
Chapter 14: Solutions Chem 103: Chapter 14
Molality Section 15.2.
Colligative Properties
COLLIGATIVE PROPERTIES
COLLIGATIVE PROPERTIES
Molality Measurement of Concentration of Solutions
Molality (m) A concentration that expresses the
Chemistry 16.4.
S-C-9-3_Concentrations Presentation
Concentration of Solute
Chapter 15 -Solubility & Concentration
Solubility & Concentration
GENERAL PROPERTIES OF SOLUTIONS
Solutions Review Chemistry.
AP Chem Take out HW to be checked Today: Solution Formation
III. Colligative Properties
Clicker #1 How many cars can we make? A) 3 B) 4 C) 6 D) 12 E) 16.
Clicker #1 How many cars can we make? A) 3 B) 4 C) 6 D) 12 E) 16.
Various Types of Solutions
BELLWORK 4/5/2018.
Chapter 18 - Solutions.
Colligative Properties
Colligative Properties of Solutions
8.3 Colligative Properties Objective S3
Calculations Involving Colligative Properties
What is the formula for molality?
Colligative Properties of Solutions
Boiling and Freezing Point Problems
Ch. 3 & 7 – The Mole II. Concentration (p )
Concentration of Solute
Clicker #1 How many cars can we make? A) 3 B) 4 C) 6 D) 12 E) 16.
Molarity Chemistry.
III. Colligative Properties (p )
Chapter 12.3 concentration –
Chapter 16 Review =“Solutions”.
Chapter 13.5 Expressing solution concentration
Pre-AP Chemistry Charles Page High School Stephen L. Cotton
Boiling and Freezing Points
Show Down 2 Step Mole Conversions
Colligative Properties of Solutions (chapter 16)
Concentration of Solution
Kinetic Molecular Theory
BELLWORK 4/6/2017.
Clicker #1 How many cars can we make? A) 3 B) 4 C) 6 D) 12 E) 16.
8.2 Measuring Concentration Objective 2
Solutions Part II.
Section 3.
Solubility & Concentration
Clicker #1 How many cars can we make? A) 3 B) 4 C) 6 D) 12 E) 16.
Colligative Properties
Calculations Involving Colligative Properties
Presentation transcript:

Step 1: Calculate the moles of sucrose. PROBLEM: You dissolve 15.0 g of sucrose, C12H22O11, in a cup of water (225 g). What is the freezing point of the solution? NOTE: Kfp (H2O) = –1.86 ºC/m Step 1: Calculate the moles of sucrose. 15.0 g C12H22O11 x 1 mol C12H22O11/ 342.3 g = 0.0438 mol C12H22O11 Step 2: Calculate the molality of C12H22O11 0.0438 mol C12H22O11 / 0.225 kg H2O = 0.195 m Step 3: Calculate DTfp DTfp = kf • m • i – – > i = 1 since sucrose is molecular solid = –1.86 ºC/m • 0.195 m = –0.362 °C Tfp = –0.362 °C