Lecture19222

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Information about Lecture19222

Published on August 10, 2008

Author: uladzimir

Source: slideshare.net

Description

a supplemental resource for students

The Properties of Mixtures: Concentration Lecture 19

Concentration is the measure of how much of a given substance there is mixed with another substance.

Concentration applies to any sort of chemical mixture, but most frequently to homogeneous solutions.

Concentration = amount or mass or volume of solute = -------------------------------------------------- amount or mass or volume of solution or solvent

Concentration = amount or mass or volume of solute = -------------------------------------------------- amount or mass or volume of solution or solvent

Molarity (M) = amount (mol) of solute ----------------------------------- volume (L) of solution

Concentration = amount or mass or volume of solute = -------------------------------------------------- amount or mass or volume of solution or solvent

Molality (m) = amount (mol) of solute ------------------------------------- mass (kg) of solvent

A sample problem on calculating molality.

Another sample problem What is the molality of a solution prepared by dissolving 339.8 g of silver nitrate in 4000. g of water? Amount of AgNO 3 =339.8/169.91 = 2.000 mol Mass of water=4.000 kg Molality of solution= 2.000/4.000=0.500 m AgNO 3

What is the molality of a solution prepared by dissolving 339.8 g of silver nitrate in 4000. g of water?

Amount of AgNO 3 =339.8/169.91 = 2.000 mol

Mass of water=4.000 kg

Molality of solution= 2.000/4.000=0.500 m AgNO 3

Another sample problem What mass of sodium phosphate is needed to prepare 0.750 m solution by dissolving it in 800. g of water? Mass of water=0.800 kg Amount of Na 3 PO 4 =0.750/0.800=0.938 mol Mass of Na 3 PO 4 = 0.938x163.94=154 g

What mass of sodium phosphate is needed to prepare 0.750 m solution by dissolving it in 800. g of water?

Mass of water=0.800 kg

Amount of Na 3 PO 4 =0.750/0.800=0.938 mol

Mass of Na 3 PO 4 = 0.938x163.94=154 g

Another sample problem What mass of water is needed to prepare 0.250 m solution by dissolving 1 kg 382 g of potassium carbonate? Mass of K 2 CO 3 =1382 g Amount of K 2 CO 3 =1382/138.21=10.00 mol Mass of water=10.00/0.250=40.0 kg

What mass of water is needed to prepare 0.250 m solution by dissolving 1 kg 382 g of potassium carbonate?

Mass of K 2 CO 3 =1382 g

Amount of K 2 CO 3 =1382/138.21=10.00 mol

Mass of water=10.00/0.250=40.0 kg

Mass percent (%w/w) = mass of solute = ---------------------------------------------- x 100% mass of solute + mass of solvent

A sample problem What is mass percent of magnesium sulfate in solution containing 43.5 g of solute in 0.348 kg of solution? Mass of solution=348 g %w/w MgSO 4 =(43.5/348) x 100%=12.5%

What is mass percent of magnesium sulfate in solution containing 43.5 g of solute in 0.348 kg of solution?

Mass of solution=348 g

%w/w MgSO 4 =(43.5/348) x 100%=12.5%

Another sample problem What is the mass of manganese (II) chloride in 640. lb of its 30.0% solution? Mass of MnCl 2 =(640x30)/100=192 lb

What is the mass of manganese (II) chloride in 640. lb of its 30.0% solution?

Mass of MnCl 2 =(640x30)/100=192 lb

Another sample problem What is the mass of 45% solution containing 9.00 kg of potassium iodide? Mass of solution=(9.00/45%)x100%=20.0kg

What is the mass of 45% solution containing 9.00 kg of potassium iodide?

Mass of solution=(9.00/45%)x100%=20.0kg

Volume percent (%v/v) = volume of solute = -------------------------- x 100% volume of solution

Mole fraction (X) = amount (mol) of solute --------------------------------------------------------- amount (mol) of solute + amount (mol) of solvent

Mole percent (mol %) = mole fraction x 100%

A sample problem What is the mole fraction and mole percent of 157.55g of nitric acid in mixture with 900.8 ml of water? Amount of HNO 3 =157.55/63.018 = 2.5000 mol Amount of H 2 O=900.8/18.016=50.00 mol Total amount = 2.5000+50.00=52.50 mol X HNO 3 =2.5000/52.50=0.04762 mol % HNO 3 = 0.04762 x 100%=4.762%

What is the mole fraction and mole percent of 157.55g of nitric acid in mixture with 900.8 ml of water?

Amount of HNO 3 =157.55/63.018 = 2.5000 mol

Amount of H 2 O=900.8/18.016=50.00 mol

Total amount = 2.5000+50.00=52.50 mol

X HNO 3 =2.5000/52.50=0.04762

mol % HNO 3 = 0.04762 x 100%=4.762%

A sample problem on expressing concentration in parts by mass, parts by volume, and mole fraction.

A sample problem on converting concentration units.

THE END

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