Methods expressing Concentrations of solutions

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Faculty of Medicine Ain Shams University Medical Biochemistry & Molecular Biology Department 1 Methods expressing Concentrations of solutions.
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Methods expressing Concentrations of solutions By All Staff Members of Medical Biochemistry Department

Apply different methods for expressing concentration By the end of this topic, the student will be able to: Apply different methods for expressing concentration

METHODS EXPRESSING THE CONCENTRATION OF A SOLUTION ? Solution Concentration ?

Solid dissolved in a Liquid. Solutions SOLVENT= material that dissolves another . SOLUTES = Substance being dissolved. Most common types of solutions involves Solid dissolved in a Liquid.

Factors determining rate of solution formation Stirring Smaller pieces. Temperature Pressure

Factors determining rate of solution formation Stirring:moves solvent into contact with the solute. Smaller pieces increase the surface area of the solute. Higher temperatures make the molecules of the solvent move faster. Pressure Factor: extra pressure will not increase solubility of Solids and Liquids (already close together) Increase pressure squeezes only gas solute into solvent.

The concentration of a solution Qualitative Expressions of Concentration Quantitative Expressions of Concentration Represents the amount of solute dissolved in a unit amount of solvent or of solution.

The amount of solute in Concentration Definite volume of solvent or solution

I. Qualitative Expressions of Concentration Diluted Concentrated

II. Quantitative Expressions of Concentration International Units of mass and volume The basic international unit of mass is the kilogram (kg) and its fractions. The basic international unit of volume is the liter (L) and its fractions .

Basic International Units MASS 1 kg = 1000 gram 1 g = 1000 milligram 1 mg = 1000 microgram 1 µg = 1000 nanogram 1 ng = 1000 picogram

Basic international units VOLUME 1 Liter (L) = 1000 mL = 10 dL 1 deciliter (dL) = 100 ml 1 milliliter (ml) = 10-3 L = 1000 µL 1 microliter (µL) = 10-6 L

Methods Of Expression Of Concentrations Percentages% Molar concentrations (M) Normality (N) Very low concentrations

Percent Concentration Describes the amount of Solute dissolved in 100 parts of solution Amount Of Solute 100 Parts Solution.

1- Percentage (%) Solvent The most commonly used two methods are according to the nature of solvent: Solvent Percent / volume Solvent Is Liquid Percent / mass Solvent Is Solid

A- Percent / volume V/V % and W/V % Further divided into 2 types according to the nature of solute: Solute volume / volume % (V/V %) Solute Is Liquid 70% ethyl alcohol (70 ml/100ml) weight / volume % (W/V %) Solute Is Solid (Commonly used) glucose 5 mg% (5mg/100ml)

B. Percent / mass (W/W) % = Mass of solute (g) / Mass of solution (g) x 100% Most common e.g. Gold Alloys (with copper or silver “Solutes”)

Percent Concentrationa %Mass/mass (سبايك الذهب و النحاس( %Volume/volume mL% Mass/volume% g %

Problem 2) 6.4% (m/m) Na2CO3 3) 6.0% (m/m) Na2CO3 A solution contains 15 g Na2CO3 and 235 g of H2O? What is the mass % of the solution? 1) 15% (m/m) Na2CO3 2) 6.4% (m/m) Na2CO3 3) 6.0% (m/m) Na2CO3

Answer Mass of solution = 15 g + 235 g = 250 g %(w/w) = 15 g Na2CO3 x 100 250 g solution = 6.0% Na2CO3 solution

Problem An IV solution is prepared by dissolving 25 g glucose in water to make total 500 mL solution. What is the (w/v %) of the glucose in the IV solution? 5.0% 20.% 50.%

Mass/volume % = (masssolute/volumesolution) 100 How much solute is present in 756.1 mL of a 14.7% (mass/volume) NaCl solution? Mass/volume % = (masssolute/volumesolution) 100 14.7 % = (x / 756.1 mL ) 100 x = 0.147 (756.1 mL) x = mass of solute = 111g

number of moles of solute in 1 L of solution 2- Molarity (M) A concentration that expresses the number of moles of solute in 1 L of solution Molarity (M) = no.moles of solute 1 liter solution (It is abbreviated as a capital M “mole/L”)

Mole The Mole is the primary SI unit of the amount of substances. Weight of 1 mole of any substance = its molecular weight (MW) in grams.

23 + (2×1) + 31 + (4 × 16)= 120 g= weight of 1 mole MW: It is the sum of the atomic weight of all atoms in the formula expressed in gram eg : one mole of NaH2PO4 23 + (2×1) + 31 + (4 × 16)= 120 g= weight of 1 mole mole The Basic International unit for matter

All solutions of the same molarity (M) in the same volume of solution= contain the same no. of solute molecules. = Avogadro´s number N = 6.022 x 1023 particles in 1 mole

In mathematical equations, the molar concentration is expressed by using the brackets [ ] 1 mole= 1000 mmole 1 millimole (mmole)= 10-3 mole 1 micromole (µmole)= 10-6 mole

Is a solution that contains one mole of a specified matter per liter. One Molar solution Is a solution that contains one mole of a specified matter per liter. 1M =1 mole/L e.g. 2 M solution of Ca =80 g/L as the molecular weight of Ca= 40g

Prepare one Molar solution of NaCl [NaCl =58.5] Weighing 58.5 g of NaCl and dissolve it in DW to final volume one litre

Prepare 0.25 M solution of NaCl Answer : 1/4 M concentration = 58.5/4 g Weighing 14.625 g of NaCl and dissolve it in DW to final volume one litre

Using 16 g NaOH, 200 ml solution is prepared. Which one of the followings is true for this solution? (NaOH = 40 g). Concentration of solution is 2 molar Volume of the water in solution is 200 ml If we add water to solution, no. Of moles of solute decrease.

Answer Calculation of Moles of NaOH: 40g NaOH = 1mole, so 16 g NaOH =?? mole = 16/40 =0.4 mole; V = 200 mL= 0.2 Liters So, Molarity = 0.4/0.2=2 molar. ………… A is true

Prepare 100 ml of 0.5 M NaCl solution. Lab activity Prepare 100 ml of 0.5 M NaCl solution. [NaCl =58.5]

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