NAME: OPTION GROUP: WJEC MATHS FOR AS BIOLOGY MAKING SOLUTIONS MATHS COVERED IN THIS BOOKLET ARE: 1. Revision of scientific notation. 2. Molar solutions. 3. Concentrations. 4. Volumes of liquids. 5. Dilutions. 6. Units of volume and concentration. 7. Significant figures, number of decimal points and rounding numbers.
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REVISION OF SCIENTIFIC NOTATION Also introducing rounding numbers, number of decimal places & significant figures. Worked examples on scientific notation 1. Express the following numbers in scientific notation. 450, 8750, 237000, 0.875, 0.000523, 12.5. 2. Express the following number as decimals. 8.9x10 1, 7.234x10-6, 0.055297x10 3, 100x10-1, 0.5x10 9. 3
Exercises on scientific notation Q1. Express the answers to the following calculations in scientific notation. 567x45, 900x2, 76 436, 0.64x0.098, 0.427 0.0994, 3.24+4, 6.3-53, Q2. State the exponent that would cause the following movement of the decimal point. (a) (b) (c) (d) The decimal point has been moved 3 places to the right. The decimal point has been moved 1 place to the left. The decimal point has been moved 4 places to the left. The decimal point has been moved 6 places to the right. Q3. What would be the exponent in the following? (a) (b) (c) (d) A tenth. A thousandth. A billion fold. A hundred fold. 4
Rounding numbers Place Values Here is the general rule for rounding numbers: 1. If the number you are rounding is followed by: 5, 6, 7,8 or 9 round the number up. 2. If the number you a rounding is followed by 0, 1, 2, 3 or 4 round the number down. What are you rounding to? When rounding a number, you first need to ask: what are you rounding it to? Numbers can be rounded to the nearest ten, hundred or thousand, and so on. Consider the number 8726 8726 rounded to the nearest ten is 8730 8726 rounded to the nearest hundred is 8800 8726 rounded to the nearest thousand is 9000 Rounding and fractions Rounding fractions works exactly the same way as rounding whole numbers. The only difference is you round to tenths, hundredths and thousandths, and so on. For example: 7.8199 rounded to the nearest tenth is 7.8. 1.0621 rounded to the nearest hundredth is 1.06 3.8792 rounded to the nearest thousandth is 3.879 5
Correcting to a specified number of decimal places Giving a number to the nearest tenth, hundredth etc. For example, 1.3839µm is 1.4µm to the nearest tenth. As tenths are represented by the first decimal point, we write: 1.3839µm = 1.4µm to 1 decimal place (1d.p.) A number given to the nearest hundredth, thousandth etc are said to be correct to 2 d.p and correct to 3 d.p respectively. So, 1.3839µm = 1.38µm to 2d.p. 1.3839µm = 1.384µm to 3d.p. Are you sure that you understand how I have/have not rounded the numbers in the above examples? Number of significant figures A mature female egg cell has a size of approximately 123µm or 123000nm. In each number, the first number (number 1) has a different place value although the number 1 is the first figure in each number. It is called the first significant figure. The figure 2 is the second significant figure. Here is the general rule for quoting a number to a certain number of significant figures. Reading any number from left to right, regardless of the decimal point, the first significant figure is the first non-zero figure. The second significant figure is the next figure, which can be a zero or any other number. You continue this for further significant figures. 6
SECTION ONE Molar solutions In this section, we will cover: 1. How to calculate the correct mass of a substance to make a molar solution of a given volume. This will involve the calculation of moles. The equations used in this section are: 7
1. Making up a molar solution. 1.1 Calculating the molar mass of glucose: The Molar Mass is just the total of the atomic masses of each element in the compound. Notes on molar mass The molar Mass for glucose is: Carbon = 12.0 g x 6 atoms = 72.0 g Hydrogen = 1.01 g x 12 atoms = 12.1 g Oxygen = 16.0 x 6 atoms = 96.0 g Total = 180.1gmol 1.2 Calculating the moles of glucose Moles is calculated using the formula: Notes on moles 8
1.3 Calculating the volume required to make a 1 molar solution of glucose You will need the following equation to do this: Notes on volume 1.4 Changing the volume and mass Notes on changing volume and mass Section 1.1-1.4 worked examples 9
Q1. Make 500cm 3 of a 1mol dm -3 solution of sodium chloride. The molar mass of sodium chloride is 58.5gmol. Q2. Make 100cm 3 of a 0.25mol dm -3 solution of maltose. The molar mass for maltose is 342gmol. Q3. Make 5dm 3 of a 2.3mol dm -3 solution of potassium chloride. The molar mass of potassium chloride is 74.5gmol. 10
Section 1.1-1.4 Exercises Q1. Make 1dm 3 of a 0.025mol dm -3 solution of calcium chloride. The molar mass of calcium chloride is 75.5gmol. Show your workings out. Q2. Make 0.3dm 3 of a 0.0025mol dm -3 solution of sodium nitrate. The molar mass of sodium nitrate is 85gmol. Show your workings out. Q3. You have 2.84g of calcium phosphate. What volume of water would you need to make a 0.004mol dm -3 solution? The molar mass of calcium phosphate is 310.18gmol. Show your workings out. 11
SECTION TWO Percentage Solutions In this section, we will cover: 1. How to make up a percentage solution. 2. Percentage Solutions There are several different percentage concentrations. There is the percentage of a solid dissolved in water this is called weight per volume (abbreviated to w/v) and the volume of a solution dissolved in water this is called volume per volume (abbreviated to v/v). Weight per volume calculation Notes on weight per volume calculations 12
Volume per volume calculation Notes on volume per volume calculations 13
Percentage solution worked examples 1. You have 15g of powdered haemoglobin. How much water would you need to make a 4% w/v solution. 2. You have 10cm 3 of ethanol. You dissolve all of this in 10cm 3 of water. What is the percentage volume per volume concentration? 3. You want a 2.3% weight per volume solution of acetone, but you only have 25cm 3 of water. What volume of acetone would you need to add to 25cm3 of water to achieve a 2.3% w/v solution? 14
Percentage solution exercises Q1 You have 100g of glucose but only 50cm 3 of water. What mass of glucose would you need to dissolve in the water to make a 25% w/v solution? Show your workings out. Q2 For an experiment you require 750cm 3 of a 5% solution of sodium chloride. How much sodium chloride would you weight out? Show your workings out. Q3 You require a 0.5% solution of ethanol for an experiment. You require a total volume of 2.5dm 3. What volume of ethanol would you measure out and what volume of water would you add it to? Show your workings out. 15
SECTION THREE How to dilute a solution using serial dilution and using the equation: V1 x C1 = V2 x C2. In this section, we will cover: 1. How to make a dilution of a solution by serial dilution and by using the equation: You can watch help videos at V1 x C1 = V2 x C2 https://thiacin.com/mathsmakingsolutionsdilutions.php 1. Serial dilutions Notes on serial dilutions 16
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Serial dilution worked examples Q1. You have a stock solution of 75moles dm -3. You require a solution of 0.075 moles dm - 3. You have 3 test tubes to do your serial dilution. Complete the diagram on page 12 by filling in the gaps to include the volumes and concentration of tubes 1 to 3. Q2. You have a stock solution of 1025moles dm -3. You require a solution of 0.001025moles dm -3. You have 3 test tubes to do your serial dilution. Complete the diagram on page 13 by filling in the gaps to include the volumes and concentration of tubes 1 to 3. 18
Diagram for worked example number 1. of stock solution added to tube 1 of tube 1 added to tube 2 of tube 2 added to tube 3 Stock Solution Tube 1 Tube 2 Tube 3 Concentration is 75 moles dm -3 Volume of water is. Volume of water is. Volume of water is. Total volume is so the dilution factor is. Total volume is so the dilution factor is. Total volume is so the dilution factor is. The concentration in this tube is now moles dm -3. i.e. 75moles dm -3. The concentration in this tube is now moles dm -3. i.e. moles dm -3. The concentration in this tube is now moles dm -3. i.e. moles dm -3. 19
Diagram for worked example number 2. of stock solution added to tube 1 of tube 1 added to tube 2 of tube 2 added to tube 3 Stock Solution Tube 1 Tube 2 Tube 3 Concentration is 1025 moles dm -3 Volume of water is. Volume of water is. Volume of water is. Total volume is so the dilution factor is. Total volume is so the dilution factor is. Total volume is so the dilution factor is. The concentration in this tube is now moles dm 3. i.e. 1025moles dm -3. The concentration in this tube is now moles dm -3. i.e. moles dm -3. The concentration in this tube is now moles dm -3. i.e. moles dm -3. 20
Serial dilution exercises Q1 Fully describe how you would dilute a 35% solution of glucose by a factor of 2. Q2 You have 1.2dm 3 of sucrose with a concentration of 0.023mol dm -3. You require 500cm 3 of a 0.00575mol dm -3 solution. Fully calculate how you would achieve this volume and concentration. Q3 You have a bottle of milk that you need to dilute by 10000 times. Fully calculate how you would achieve this. 21
2. Using V1 x C1 = V2 x C2 The above equation allows you to calculate the concentration of a solution and the exact volume you require from your stock solution. V1 and C1 are the concentration and volume of your stock solution and V2 and C2 are the concentration and volume of the solution that you require. Notes on V1 x C1 = V2 x C2 22
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Worked examples on V1 x C1 = V2 x C2 Q1. You have 500cm 3 of stock solution of concentration 56mol dm -3. For an experiment you require 175cm 3 of sucrose at a concentration of 12.5mol dm -3. Calculate the volume of stock sucrose and water required. Q2. You have diluted a stock solution of glucose to 76mol dm -3. The total volume of your diluted stock solution was 250cm 3. Your stock solution volume was 35cm 3. Calculate the concentration of the stock solution. Q3. Your stock solution has a concentration of 34%. You add 36cm 3 of this stock solution to 200cm 3 of water. What is the concentration of the diluted solution? 24
Exercises on V1 x C1 = V2 x C2 Q1. You require a 50cm 3 solution of potassium chloride at a concentration of 12.5mol dm 3. You have a stock solution of concentration 62mol dm -3. Calculate the volume of stock solution you require. Q2 You have been given a 6.2mol dm -3 solution of glucose. You are told that this has been made by adding 23.2cm 3 of the stock solution to 5.6cm 3 of water. Calculate the concentration of the stock solution. Q3. You have a stock solution of concentration 84%. Your working solution concentration is 72.6%. You used 26cm 3 of the stock solution. Calculate the volume of your working solution. 25