post lab 2 - Question 1­4 Parallel Dilution Trial Serial...

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Unformatted text preview: Question 1­4 Parallel Dilution Trial Serial Dilution Trial Both Dilution Trial Question 4 a b c d 1 2 3 4 5 6 7 Concentration of KMnO Absorbance for Parallel 1.00 1.760 0.60 1.360 0.40 1.015 0.20 0.534 0.10 0.247 0.05 0.129 0.02 0.053 1 2 3 4 5 6 7 8 Concentration of KMnO Absorbance for Serial 2.000 1.900 1.000 0.940 0.500 0.478 0.250 0.241 0.125 0.124 0.063 0.079 0.031 0.037 0.016 0.022 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Concentration of KMnO Absorbance 1.00 1.760 0.60 1.360 0.40 1.015 0.20 0.534 0.10 0.247 0.05 0.129 0.02 0.053 2.000 1.900 1.000 0.940 0.500 0.478 0.250 0.241 0.125 0.124 0.063 0.079 0.031 0.037 0.016 0.022 yes no Yes, the parallel dilution best fit line is better then the serial. It shows that the parallel technique is m Their solutions could have been more concentrated and not diluted correctly, that's why the data p Question 5 Both Dilution Trial 1 2 3 4 5 6 9 10 Concentration of KMnO4 0.60 0.40 0.20 0.10 0.05 0.02 0.250 0.125 11 12 13 0.063 0.031 0.016 Question 6 y=2.3161x ­ 0.0413 0.258 = y 0.258 = 2.3161x­0.0413 x = 0.129225 Question 7 The formula weight of NaCl is 58.4g we need to prepare a 900 mL of 0.5 M NaCl. First, we have to so the weight of NaCl needed is _________. Once we have weighed out the required mass of NaC Question 8 To prepare a 30 mL of a 0.5 M NaCl from a 2M NaCl stock solution. I will first find the intial volume After I have calculated the volume of V1, then ill use the appropriate measuring device to transfer t Question 9 To prepare the solution again, I will find the initial volume using the formula c1v1=c2v2 and c1= 2M then solve for v1 which is 0.225mL which is equalient to 225 Question 10 Before we go ahead with calculatuion, we have to test if the dilution factor is constant. If the dilution 2m*10ml / 2m = 10ml v1 = 1.5*10ml / 2m = 7.5ml v1= 1m *10ml /2m = 5ml solution 2M 1.5M 1M volume of 2M nacl sln needed 10mL 7.5mL 5mL First I label 3 test tubes, then I will use an appropriate measuring device to transfer the required amount of 2M question 11 Question 12 since 1 mL of water weigh 1g therefore 900mL will also weigh about 900g. A 900mL is equal to 900 a The student should have poured in half of the volume of water he or she needed instead of pouring b the concentration = number of solute/amount of solution change in mL into L by dividing 300mL by c percentage error = (0.227­0.25) / .25*100 = ­0.023/0.25 * 100 = 9.2 % Concentration changes of KMnO4 using Parallel Dilutio 2.000 Absorbance 1.500 f(x) = 1.81x + 0.12 R² = 0.96 1.000 Absorbance for Parallel 0.500 0.000 0.00 0.20 0.40 0.60 Linear Regres sion for Absorbance for 0.80 1.00 1.20 Concentration of KMnO4 (mM) Concentration changes of KMnO4 using Serial Dilutio 2.000 Absorbance f(x) = 0.94x + 0.01 1.500 R² = 1 1.000 Absorbance for Serial 0.500 0.000 0.000 0.500 1.000 Linear Regres sion for Ab1.500 sorbance for 2.000 2.500 Concentration of KMnO4 (mM) tter then the serial. It shows that the parallel technique is more precise out of the two techniques used. ncentrated and not diluted correctly, that's why the data points would not be linear. Also we know that the closer the data is to the values o PLOT GRAPH 1.360 1.015 0.534 0.247 0.129 0.053 0.241 0.124 Concentration of KMnO4 and Absorbance using both Di 1.500 Absorance Absorbance 1.000 f(x) = 2.32x - 0.04 R² = 0.93 0.500 Absorbance 0.000 Linear Regre for Absorban 0.079 0.037 0.022 Absorance 1.500 1.000 f(x) = 2.32x - 0.04 R² = 0.93 0.500 0.000 0.00 Absorbance 0.10 0.20 0.30 0.40 Linear Regre for Absorban 0.50 0.60 Parallel and Serial Dilution concentration (mM) this is the concentration of the unknown solution. eed to prepare a 900 mL of 0.5 M NaCl. First, we have to convert mL to liters by dividing 900 mL by 1000 giving 0.9L. Molarity is given in __. Once we have weighed out the required mass of NaCl needed, using a top loading balance and a small weigh boat, then we mis abou a 2M NaCl stock solution. I will first find the intial volume using the formula: C1V1 = C2V2, C1 = 2M, C2=0.5M, V2 = 30mL, V1 = ? So 0.5 then ill use the appropriate measuring device to transfer the 7.5 mL solution into a 30 mL graduated cylinder. Ill make sure I add enough d he initial volume using the formula c1v1=c2v2 and c1= 2M, c2 = 0.005M, v2 = 90ml and v1=? is equalient to 225 µL. Ill transwfer this amount to a graduated cylinder usign a pipette. Then I will atttach a piece of parafilm and gently in have to test if the dilution factor is constant. If the dilution factor is constant then we have to use serial dilution technique, however in this volume of dH2o needed (mL) 0mL 2.5mL 5mL e measuring device to transfer the required amount of 2M of sln into a 10 mL g. cylinder. Then I will add enough dH2o to bring the total vo 00mL will also weigh about 900g. A 900mL is equal to 900g of the same solution. So 0.02% thimerasol is 0.02/100*900g = 0.18g of thimer f the volume of water he or she needed instead of pouring the wholoe 300mL of water. The student should have poured in approx 150 mL ount of solution change in mL into L by dividing 300mL by 1000 so you end up with 0.3 L and for the 330mL you will end up with 0.33 L. th = ­0.023/0.25 * 100 = 9.2 % arallel Dilution 2.000 ce for gresbfor 1.00 1.800 f(x) = 1.86x - 0.09 R² = 0.97 1.600 1.400 1.20 Absorbance for Parallel 1.200 1.000 Absorbance for Serial 0.800 Linear Regres sion for Absorbance for Serial Regres Linear 0.600 0.400 Serial Dilution 0.200 0.000 0.00 0.20 0.40 2.000 1.800 ce for 1.600 gresbfor 0 0.60 0.80sion for Ab1.00 sorbance for Serial f(x) = 1.81x + 0.12 f(x) = 1.86x - 0.09 R² R²0.96 = = 0.97 1.400 1.200 Absorbance for Parallel 0.600 Linear Regres sion for Absorbance for Parallel Absorbance for 0.400 2.500 1.20 Serial 1.000 0.800 0.200 Linear Regres 0.000 sion for Ab0.00 0.20 0.40 0.60 0.80 1.00 sorbance for e data is to the values of 1 or ­1 the more closer the scatter point will form a straight line. sing both Dilution Techniques Absorbance Linear Regression for Absorbance 1.20 Absorbance Linear Regression for Absorbance 0.50 0.60 0.70 entration (mM) 0.9L. Molarity is given in moles per 1 Liter. Molarity is 0.5 M. The weight of NaCl sample that we need to calculate can be calculated by m boat, then we mis about 450mL of dH2O into a beaker and place the beaker on a stir plate. Then gently pout the wieghted amount of Na 2 = 30mL, V1 = ? So 0.5M*30mL / 2M = 7.5 mL. ake sure I add enough dH2O to bring the total volume to 30mL. Then I will attach a piece of parafilm over the graduated cylinder and then of parafilm and gently invert it several times. Finally I will label the g. cylinder 5mM. chnique, however in this case the factor is not constant therefore we have to use parallel techinque. C1V1=C2V2, c1 = 2m, v1 = ?, v2 = 10 H2o to bring the total volume to 10mL. Then I will attach a piece of parafilm over the top of the graduated cylinder. and then mix the conte 0*900g = 0.18g of thimerasol. To prepare the solution I would weigh out 0.18g of thimerasol and dissolve in less than 900mL H2Oof water oured in approx 150 mL because the weight of the sucrose will increase the volume of the solution in the beaker. ill end up with 0.33 L. then you 0.25M * 0.30L / 0.33L = 0.075/0.33 = 0.227M can be calculated by multiplying the formula weight and volume by using hte equation weight of solute (g) = formula weight of solute (g/m e wieghted amount of NaCl slowly into the beaker and stirring gently. Once the solution is dissolved then I will transfer it inside a graduated duated cylinder and then gently invert it several times. finally i will attach a peice of tape that reads 0.5 M on the measuring device. , c1 = 2m, v1 = ?, v2 = 10mL, c2 ­ 2m, c2 = 1.5 M c2 = 1M. using the formula, r. and then mix the content. i will pout the contents into the test tube named NaCl then i just prepared. then ill rinse my graduated cylinde han 900mL H2Oof water then add enough water to procduce a final volume of 900mL. then i will attach a peice of parafilm and gently seal ula weight of solute (g/mole) X molarity (mol/L) X final volume (L) sfer it inside a graduated cylinder and seal it with a parafilm around the top and gently inverting the the cylinder a few times until all the so measuring device. se my graduated cylinder and prepare the rest of my dilution the same way. f parafilm and gently seal and invert it then label the cylinder 0.02% thimerasil. few times until all the solute has been dissolved. I will then use a mesuaring device to bring the volume up to 900mL. THen i will attach a 0mL. THen i will attach a peice of tape that reads 0.5 M solution of NaCl. ...
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