notes13and14 - Notes chapters 13 and 14 Chapter 13- Notes...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Notes chapters 13 and 14 Chapter 13- Notes Terms: Solvent: component doing the dissolving and whose state is maintained. ( state meaning solid, liquid, or gas. ) Solute: being dissolved and whose state is changed. Two way of expressing the concentration of a solution are (a) Molarity or M for short, and (b) % Mass. Molarity (M) Molarity = moles of solute liters of solution Example: if 45 g of NaOH is placed in enough water to make 375 ml of a solution, what is its molarity? M = moles of solute/ liters of solution First , we must change 45 g of NaOH to moles: (45g NaOH/1)(1 mole NaOH/ 40 g NaOH) = 1.1 moles Second , the volume must be in liters. (375 ml/1)( 1L / 1000 ml) = .375 L Now substitute into the formula: M = moles/L = 1.1 moles / .375 L = 2.9 M The Molarity equation can be rearranged algebraically to calculate for moles and Liters. M = moles / L moles = M X L L = moles / M
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Example problem: 3 L of 4.5 M NaCl contain how many moles of NaCl? moles = M X L = 4.5 M X 3 L = 13.5 moles of NaCl. To calculate grams of NaCl: (13.5 moles NaCl / 1)( 58.5 g NaCl / mole) = 789.8 g % Mass Problems To calculate the % mass of a component in a solution, use the general math
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 01/03/2011 for the course CHEM 3302 taught by Professor Dickpowell during the Spring '10 term at UT Arlington.

Page1 / 5

notes13and14 - Notes chapters 13 and 14 Chapter 13- Notes...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online