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Protocol: This exercise is to be completed individually. Record the actual values for masses, volumes, etc. to create the various solutions as these will be needed for the assignment. Preparation of Stock Solutions (1) Prepare approximately 1 mL of 30.0% w/v NaCl from solid NaCl in a microcentrifuge tube. (2) Prepare approximately 1 mL of 4.0 M NaCl from solid NaCl in a microcentrifuge tube. (3) Prepare 1 mL of 25.0% v/v ethylene glycol in d$^2$H$_2$O in a microcentrifuge tube. (4) Prepare 1 mL of 2.0 M ethylene glycol ($\rho_{ethylene\ glycol}$=1.115 g/mL) in microcentrifuge tube. You will be provided a 0.5 M acetate buffer at pH 4.8 in a stock bottle and a 50 mg/mL (highly concentrated) solution of lysozyme. (5) Using your freshly created stock solutions, prepare the solutions listed in Table 3 in separate 'reservoir' areas of a 24 well crystallization plate. 

          Protocol:
This exercise is to be completed individually. Record the actual values for masses, volumes, etc. to
create the various solutions as these will be needed for the assignment.
Preparation of Stock Solutions
(1) Prepare approximately 1 mL of 30.0% w/v NaCl from solid NaCl in a microcentrifuge tube.
(2) Prepare approximately 1 mL of 4.0 M NaCl from solid NaCl in a microcentrifuge tube.
(3) Prepare 1 mL of 25.0% v/v ethylene glycol in d$^2$H$_2$O in a microcentrifuge tube.
(4) Prepare 1 mL of 2.0 M ethylene glycol ($\rho_{ethylene\ glycol}$=1.115 g/mL) in microcentrifuge tube.
You will be provided a 0.5 M acetate buffer at pH 4.8 in a stock bottle and a 50 mg/mL (highly
concentrated) solution of lysozyme.
(5) Using your freshly created stock solutions, prepare the solutions listed in Table 3 in separate
'reservoir' areas of a 24 well crystallization plate.
Show more…
Protocol: This exercise is to be completed individually. Record the actual values for masses, volumes, etc. to create the various solutions as these will be needed for the assignment. Preparation of Stock Solutions (1) Prepare approximately 1 mL of 30.0% w/v NaCl from solid NaCl in a microcentrifuge tube. (2) Prepare approximately 1 mL of 4.0 M NaCl from solid NaCl in a microcentrifuge tube. (3) Prepare 1 mL of 25.0% v/v ethylene glycol in d$^2$H$2$O in a microcentrifuge tube. (4) Prepare 1 mL of 2.0 M ethylene glycol (ρethylene glycol=1.115 g/mL) in microcentrifuge tube. You will be provided a 0.5 M acetate buffer at pH 4.8 in a stock bottle and a 50 mg/mL (highly concentrated) solution of lysozyme. (5) Using your freshly created stock solutions, prepare the solutions listed in Table 3 in separate 'reservoir' areas of a 24 well crystallization plate.

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Biology for AP Courses
Biology for AP Courses
Julianne Zedalis, John Eggebrecht
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Protocol: This exercise is to be completed individually. Record the actual values for masses, volumes, etc. to create the various solutions as these will be needed for the assignment. Preparation of Stock Solutions (1) Prepare approximately 1mL of 30.0%(w)/(v)NaCl from solid NaCl in a microcentrifuge tube. (2) Prepare approximately 1mL of 4.0MNaCl from solid NaCl in a microcentrifuge tube. (3) Prepare 1mL of 25.0%(v)/(v) ethylene glycol in d^(2)H_(2)O in a microcentrifuge tube. (4) Prepare 1mL of 2.0M ethylene glycol ( ho _(ethylene glycol )=1.115(g)/(m)L ) in microcentrifuge tube. You will be provided a 0.5M acetate buffer at pH4.8 in a stock bottle and a 50m(g)/(m)L (highly concentrated) solution of lysozyme. (5) Using your freshly created stock solutions, prepare the solutions listed in Table 3 in separate 'reservoir' areas of a 24 well crystallization plate. Protocol: This exercise is to be completed individually. Record the actual values for masses, volumes, etc. to create the various solutions as these will be needed for the assignment. Preparation of Stock Solutions (1) Prepare approximately 1 mL of 30.0% w/v NaCl from solid NaCl in a microcentrifuge tube. (2) Prepare approximately 1 mL of 4.0 M NaCl from solid NaCl in a microcentrifuge tube. (3) Prepare 1 mL of 25.0% v/v ethylene glycol in d2H,O in a microcentrifuge tube. You will be provided a 0.5 M acetate buffer at pH 4.8 in a stock bottle and a 50 mg/mL (highly concentrated) solution of lysozyme. (5) Using your freshly created stock solutions, prepare the solutions listed in Table 3 in separate 'reservoir' areas of a 24 well crystallization plate.
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Transcript

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00:01 Hello students in this video going to study about the titration.
00:02 For that we have molarity of c a x ion in the standard solution and that is number of moles into c .o .3 and that is again mass into cco3 and this is molar mass.
00:13 Let's solve it for that we have 0 .114 gram divided by 100 .09 gram per mole and then we solve it we get 0 .0000, double 1 .0 3 mole and for that again we have m ca2 plus please do not get confused this ca2 plus stands for calcium is equals to and this is the mass and ca2 plus divided by volume and that we get it as 0 .00113 mole divided by 0 .250 letter and then when we solve it we get 0 .00004 1 to m and now we are going to find out the molar concentration of edta in solution for that we have molar concentration of edta in solution for that we have nedta is equal to n c a plus that is again is equal to m c a 2 plus into v c c ca 2 plus and now we have put in the values forwarding at the values we get 0 .0021 mole and medta is nothing but medta is equals to 0 .0021 mole divided by 0 .02062 litter and then when you solve it we get 0 .0 .01062 litter and then when you solve it we get 0 .0 .0 .07062 litter and then when you solve it we get 0 .0 .07 m and now the concentration of the concentration the concentration the concentration of edta solution expressed as ca this is that is calcium so what we do we get one mole of ca divided by one mole edta into 40 .08 c a divided by one mole of c a now that into 1000mg divided by one gram and that is again divided by 0 .0606 to letter when we solve it what do we get solving it we get one nine four six millicram of calcium per milliliter and now we have the amount of ca in 50 ml and the mass of ca in 50 ml.
03:45 For that we have 0 .004412m into 0 .0 .050 liter.
03:58 For that we have 0 .3 .0 .2 mole.
04:05 This is the mole and now when we are converting in grams we have 0 .00 to 1 mole into 40 .08 gram per mole and when we solve it we get 8 .86mg of ca and now we are going to find out the ml of 0 .010m e dta solution needed solution needed to titrate the sample for that we have e .d .a...
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