ALL work must be shown . Please show me detailexplanation ( thanks ) Kalam(MNK)

Question

ALL work must be shown . Please show me detailexplanation ( thanks ) Kalam(MNK) 1. Assume that you have only 5 ml of expensive diluentand the smallest volume that your pipets can dispense is 10 ul. Youhave other pipets that can deliver 0.1-10 ml and various graduatedcylinders or volumetric fla1sks that can dispense largervolumes. Describe serial dilutions that you can perform todilute a 0.7 M solution with the valuable diluent to make a 0.7 nMsolution. You need only 1 ml of the final 0.7 nM solution. 2. Assume that you need to dilute a 5 mM solution to aconcentration of 1 x 10 –5 M. Pipettors thatdispense volumes ranging from 10 -100 µl and 100 - 1000µl volumes and volumetric flasks for measuring 10 ml or 100ml are available. Only 1.5 mL of diluent is available. Youneed only 1000 µl of the final solution     ALL work must be shown . Please show me detailexplanation ( thanks ) Kalam(MNK) 1. Assume that you have only 5 ml of expensive diluentand the smallest volume that your pipets can dispense is 10 ul. Youhave other pipets that can deliver 0.1-10 ml and various graduatedcylinders or volumetric fla1sks that can dispense largervolumes. Describe serial dilutions that you can perform todilute a 0.7 M solution with the valuable diluent to make a 0.7 nMsolution. You need only 1 ml of the final 0.7 nM solution. 2. Assume that you need to dilute a 5 mM solution to aconcentration of 1 x 10 –5 M. Pipettors thatdispense volumes ranging from 10 -100 µl and 100 - 1000µl volumes and volumetric flasks for measuring 10 ml or 100ml are available. Only 1.5 mL of diluent is available. Youneed only 1000 µl of the final solution 2. Assume that you need to dilute a 5 mM solution to aconcentration of 1 x 10 –5 M. Pipettors thatdispense volumes ranging from 10 -100 µl and 100 - 1000µl volumes and volumetric flasks for measuring 10 ml or 100ml are available. Only 1.5 mL of diluent is available. Youneed only 1000 µl of the final solution    

Explanation / Answer

smallest volume that your pipets can dispense is 10l Concentration of stock solution , M = 0.7 M Concentration of required solution , M' = 0.7 nM = 0.7 * 10^-9M Volume of the solution required , V' = 1 mL Volume of stock taken , V = ? According to dilution law MV = M'V'                                           V = 0.7 * 10^-9 M * 1 mL / 0.7 M                                               = 10^-9 mL                                                =10^-9 * 10^-3 L                                               = 10^-12 L                                               = 10^-6 * 10^-6 L                                                =10^-6L                            Since 1L = 10^-6 L but the smallest volume that your pipets can dispense is 10l , which is impossible by direct dilutio, so we have tofollow a series of dilutions to acquire the requiredsolution. 1 mL = 10^-3 L          = 10^-3* 10^-6 * 10^6 L          = 10^3 *10^-6 L          = 1000L So , 5 mL of the soltion is = 5 * 1000L                                        = 5000 L We have only 5000L of the soltion for dilution. Follow the series :- ----------------- (1) MV = M'V'     0.7 M * V = 0.007 M * 1000L                 V = 10 L take 10L of the stock & diluted to 1000L =1mL,So 1mL is consumed,4mL is remaining (2) MV = M'V'    0.007 M * V = 0.00007 M * 1000 L                     V = 10 L take 10 L of the above diluted solution is taken &diluted to 1000L = 1 mL. So 1mL + 1mL = 2 mL is consumed&3 mL is remained. (3) MV = M'V'    0.00007 M * V = 7 * 10^-7M * 1000 L                     V = 10 L take 10 L of the above diluted solution is taken &diluted to 1000L = 1 mL. So 2mL + 1mL = 3 mL isconsumed& 2 mL is remained. (3) MV = M'V'    0.00007 M * V = 7 * 10^-7M * 1000 L                     V = 10 L take 10 L of the above diluted solution is taken &diluted to 1000L = 1 mL. So 2mL + 1mL = 3 mL isconsumed& 2 mL is remained. (4) MV = M'V' 7*10^-7 M * V = 7*10^-9M * 1000 L                     V = 10 L take 10 L of the above diluted solution is taken &diluted to 1000L = 1 mL. So 3mL + 1mL = 4mL is consumed&1mL is remained. (5) MV = M'V' 7*10^-9 M *V = 7*10^-10M * 100 L                     V = 10 L take 10 L of the above diluted solution is taken &diluted to 100L = 0.1 mL. So 4mL + 0.1mL = 4.1 mL is consumed& 0.9 mL is remained. (4) MV = M'V' 7*10^-7 M * V = 7*10^-9M * 1000 L                     V = 10 L take 10 L of the above diluted solution is taken &diluted to 1000L = 1 mL. So 3mL + 1mL = 4mL is consumed&1mL is remained. (5) MV = M'V' 7*10^-9 M *V = 7*10^-10M * 100 L                     V = 10 L take 10 L of the above diluted solution is taken &diluted to 100L = 0.1 mL. So 4mL + 0.1mL = 4.1 mL is consumed& 0.9 mL is remained. (5) MV = M'V' 7*10^-9 M *V = 7*10^-10M * 100 L                     V = 10 L take 10 L of the above diluted solution is taken &diluted to 100L = 0.1 mL. So 4mL + 0.1mL = 4.1 mL is consumed& 0.9 mL is remained. So we can prepare the required solution of 7 * 10^-10 L = 0.7* 10^-9 L = 0.7 nM solution     Simillarly do the second one     

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