Hey guys, can u please help me on Q1 (a) ii.
http://www.xtremepapers.com/CIE/International A And AS Level/9700 - Biology/9700_w03_qp_3.pdf
http://www.xtremepapers.com/CIE/International A And AS Level/9700 - Biology/9700_w03_qp_3.pdf
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Preparing sol. bio p3
- Thread starter TSZ
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To prepare 0.5 % glucose solution from 5% glucose solution, we carry out simple dilution. The formula for simple dilution is M1V1=M2V2 where M1=% conc. of solution which you have to dilute (in this case 5%) V1=volume of the (5% glucose) solution used to make 0.5% glucose solution, M2=% conc. of dilute solution (in this case 0.5%), V2=volume of dilute solution prepared (which is the sum of V1 and volume of distilled water added).
They would have given you atleast 10cm3 of 5% sol., a beaker with distilled water and two 10cm3 syringes. Usually they tell you the volume of dilute solution which you have to prepare (V2). As they havent told you in this paper, you can decide for yourself. As they have given you 10cm3 of 5% sol. you can prepare 10cm3 of 0.5% sol. so V2=10cm3. M1 is 5% and M2 is 0.5%. By substituting these values into M1V1=M2V2 we get V1=1 cm3. This is the volume of 5% sol. that you will use. Since, you have to prepare 10cm3 of 0.5% solution the volume of distilled water used will be= 10 - V1 = 10 - 1 = 9cm3.
In Examination you will write:- Add 9 cm3 of distilled water to 1 cm3 of 5% sol. to get 10cm3 of 0.5% solution.
To prepare 0.25%, we will use the 0.5% solution (which we prepared with simple dilution) and distilled water. Now we will carry out serial dilution (not simple dilution), as we have to halve the conc. (0.5% to 0.25%), so we simply add equal volumes of 0.5% sol. and distilled water to prepare 0.25% sol. For example, add 5 cm3 of 0.5% sol. and 5 cm3 of distilled water to get 10 cm3 of 0.25% sol.
In Examination you will write:- Add 5 cm3 of distilled water to 5 cm3 (equal volume) of 0.5% sol. to get 0.25% sol.
Hope its not too confusing!
btw...if you have problem understanding the theory of it, i can use a diagram to explain it!
They would have given you atleast 10cm3 of 5% sol., a beaker with distilled water and two 10cm3 syringes. Usually they tell you the volume of dilute solution which you have to prepare (V2). As they havent told you in this paper, you can decide for yourself. As they have given you 10cm3 of 5% sol. you can prepare 10cm3 of 0.5% sol. so V2=10cm3. M1 is 5% and M2 is 0.5%. By substituting these values into M1V1=M2V2 we get V1=1 cm3. This is the volume of 5% sol. that you will use. Since, you have to prepare 10cm3 of 0.5% solution the volume of distilled water used will be= 10 - V1 = 10 - 1 = 9cm3.
In Examination you will write:- Add 9 cm3 of distilled water to 1 cm3 of 5% sol. to get 10cm3 of 0.5% solution.
To prepare 0.25%, we will use the 0.5% solution (which we prepared with simple dilution) and distilled water. Now we will carry out serial dilution (not simple dilution), as we have to halve the conc. (0.5% to 0.25%), so we simply add equal volumes of 0.5% sol. and distilled water to prepare 0.25% sol. For example, add 5 cm3 of 0.5% sol. and 5 cm3 of distilled water to get 10 cm3 of 0.25% sol.
In Examination you will write:- Add 5 cm3 of distilled water to 5 cm3 (equal volume) of 0.5% sol. to get 0.25% sol.
Hope its not too confusing!
btw...if you have problem understanding the theory of it, i can use a diagram to explain it! 
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WOW.... i understood it.. GREAT!!
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UM quick question.. Y DOES THE MS SAY make up till 2??????????????? plz help