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**II. Solution Concentration (p. 480 – 486)**

Ch. 16 – Solutions II. Solution Concentration (p. 480 – 486)

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**A. Concentration The amount of solute in a solution**

Describing Concentration % by mass - medicated creams % by volume - rubbing alcohol ppm, ppb - water contaminants molarity - used by chemists molality - used by chemists

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**substance being dissolved**

B. Percent Solutions Percent By Volume (%(v/v)) Concentration of a solution when both solute and solvent are liquids often expressed as percent by volume substance being dissolved total combined volume

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B. Percent Solutions Find the percent by volume of ethanol (C2H6O) in a 250 mL solution containing 85 mL ethanol. Solute = 85 mL ethanol Solution = 250 mL 85 mL ethanol 250 mL solution % (v/v) = = 34% ethanol (v/v) x 100

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**substance being dissolved**

B. Percent Solutions Percent By Mass (%(m/m)) Concentration of a solution when solute is solid sometimes expressed as percent by mass substance being dissolved total combined volume

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B. Percent Solutions How much glucose should you use to make g of a 2.8% (m/m) solution in water? Percent (m/m) = 2.8% Solution = g unknown g glucose 2000. g solution 2.8 = = g glucose x 100

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**substance being dissolved**

C. Molarity Concentration of a solution most often used by chemists substance being dissolved total combined volume

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C. Molarity 2M HCl What does this mean?

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**D. Molarity Calculations**

LITERS OF GAS AT STP Molar Volume (22.4 L/mol) MASS IN GRAMS MOLES NUMBER OF PARTICLES Molar Mass (g/mol) 6.02 1023 particles/mol Molarity (mol/L) LITERS OF SOLUTION

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**D. Molarity Calculations**

How many moles of NaCl are required to make 0.500L of 0.25M NaCl? 0.500 L sol’n 0.25 mol NaCl 1 L sol’n = mol NaCl

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**D. Molarity Calculations**

How many grams of NaCl are required to make 0.500L of 0.25M NaCl? 0.500 L sol’n 58.44 g NaCl 1 mol NaCl 0.25 mol NaCl 1 L sol’n = 7.3 g NaCl

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**D. Molarity Calculations**

Find the molarity of a 250 mL solution containing 10.0 g of NaF. 10.0 g NaF 1 mol NaF 41.99 g NaF = 0.95 M NaF .25 L sol’n

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E. Dilution Preparation of a desired solution by adding water to a concentrate Moles of solute remain the same

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E. Dilution What volume of 15.8M HNO3 is required to make 250 mL of a 6.0M solution? GIVEN: M1 = 15.8M V1 = ? M2 = 6.0M V2 = 250 mL WORK: M1 V1 = M2 V2 (15.8M) V1 = (6.0M)(250mL) V1 = 95 mL of 15.8M HNO3

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F. Molality mass of solvent only 1 kg water = 1 L water

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**G. Molality Calculations**

Find the molality of a solution containing 75 g of MgCl2 in 250 mL of water. 75 g MgCl2 1 mol MgCl2 95.21 g MgCl2 0.25 kg water = 3.2m MgCl2

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**G. Molality Calculations**

How many grams of NaCl are req’d to make a 1.54m solution using kg of water? 0.500 kg water 1.54 mol NaCl 1 kg water 58.44 g NaCl 1 mol NaCl = 45.0 g NaCl

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**H. Preparing Solutions 1.54m NaCl in 0.500 kg of water**

500 mL of 1.54M NaCl mass 45.0 g of NaCl add water until total volume is 500 mL mass 45.0 g of NaCl add kg of water 500 mL water 45.0 g NaCl 500 mL mark volumetric flask

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H. Preparing Solutions Copyright © NT Curriculum Project, UW-Madison (above: “Filling the volumetric flask”)

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H. Preparing Solutions Copyright © NT Curriculum Project, UW-Madison (above: “Using your hand as a stopper”)

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**H. Preparing Solutions 250 mL of 6.0M HNO3 by dilution**

measure 95 mL of 15.8M HNO3 95 mL of 15.8M HNO3 combine with water until total volume is 250 mL 250 mL mark Safety: “Do as you oughtta, add the acid to the watta!” or AA – add acid! water for safety

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**Solution Preparation Mini-Lab**

Turn in one paper per team. Complete the following steps: A) Show the necessary calculations. B) Write out directions for preparing the solution. C) Prepare the solution. For each of the following solutions: 1) mL of 0.50M NaCl 2) 0.25m NaCl in mL of water 3) mL of 3.0M HCl from 12.0M concentrate. (don’t actually prepare this one!)

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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 -

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 -

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