1. Given the following serial dilution: Volume Transferred: 2 ml 1 ml 5 ml 3 ml
ID: 198187 • Letter: 1
Question
1. Given the following serial dilution: Volume Transferred: 2 ml 1 ml 5 ml 3 ml Initial Tube Final Tube Volume of Diluent: 10 m 5 ml 4 ml 20 ml a. What is the concentration in the final tube if the concentration in the initial tube was What is the concentration in the initial tube if the concentration in the final tube was 15uM7 C1, l= b. 2. Show how you would construct a serial dilution if you were given a 1M solution and you need exactly 10mls of a 20 nanomolar solution. 3. How many microliters of a stock solution of protein that is 21mg/ml would you transfer to get 6mls of a 0.04mg/ml solution? 4. You are given 1ml of a 30mM stock solution of NADPH. You are told to dilute it so that its absorbance in the Spec20 equals 0.60. The molar extinction coefficient for NADPH is 7105/mole cm and the path length is 1.6cm. The volume of the tube is 10mls. How many microliters of stock NADPH do you transfer? 5. As the lab instructor for Cell Biology, you need to make a solution of DCMU, an inhibitor of photosynthetic electron transport. You want a final concentration of 2x104M in a final volume of 2mls, and this needs to be achieved using a 10ul shot from the micropipette. How much DCMU should you weigh out to dissolve in 10mls of alcohol? The MW of DCMU is 400g/mole.Explanation / Answer
In this question final volume has non given.
Final volume in second tube is = 10 +1= 11 ml
4.2* 1 = C2 * 11
4.1*1/11= 0.38 M
In second final volume is 10 ml
0.38 * 5= C3 * 10
0.19 =C3
0.19*2/6 =0.063= C4
0.063 * 3 = 23 C5
0.0083 M= C5
3* C4= 23 * 15
C4= 115
C3 *2 = 115*6
C3= 345
C2* 5= 345*10
C2= 690
C1 *1 = 690 *11
C1= 7590
2. For this we do the serial dilution
1ml of 1M will be added in 9 ml so it become the 0.1 M
1ml of 0.1M will be adde to 9 ml so it become 0.01M
Than similar lu 0.001 or 10-3 M or 1mM
Similar next dilution 10-4 M
Similar next dilution 10-5 M
Similar next dilution 10-6 M or 1 micro molar
Similarly 10-7 M
X * 10-7 M = 20 10-9 M * 10
X= 200 * 10-2 ml
X= 2ml
And remain 18 ml of water
Here 2 ml o 10-7 M solution and 18 ml of water is required.
3. Here 1ml = 1000ul
21 mg/ml * X = 0.04 * 6
X= 0.01142 ml
Or
0.01142 or 11.42 ul and final volume is 6 mls
5.
First you need find our the Molarity in 10 ul
2 * 10-4 M * 2000 ul= 10 * X
400 *10-4 M = X
Now according to Molarity formula
400 *10-4 M= wt in gm/ mol weigh * volume ( lit)
(400 10-4 * 400*10)/1000= Wt. in grams
40/1000= 40 mg is required to dissolve in 10 ml
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