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Answers & Comments
nicoleannseli
Concentration Questions & Answers How many grams of glucose would be dissolved to make 1 liter of a 0.5M glucose solution? 0.5 mole/liter x 180 grams/mole x 1 liter = 90 g How many molecules of glucose are in that 1 liter of 0.5M glucose solution? 0.5 mole/liter x 1 liter x 6.023x1023 molecules/mole = 3.012x1023 molecules What is the concentration of the 0.5M glucose solution expressed in mM? 0.5 mole/liter x 1000 mmoles/mole = 500 mM What is the concentration of the 0.5M glucose solution expressed in %? 0.5 mole/liter x 180 grams/mole x 1 liter/1000ml = 0.09 g/ml = 9 g/100ml = 9% How many grams of sucrose would be dissolved in 1 liter of a 0.5M sucrose solution? How does that compare to the grams of solute in the 0.5M glucose solution? 0.5 mole/liter x 342 grams/mole x 1 liter = 171 g = almost twice as many grams How many molecules of sucrose in that 1 liter of 0.5M sucrose solution? How does that compare to the amount of solute in the 0.5M glucose solution? 0.5 mole/liter x 1 liter x 6.023x1023 molecules/mole = 3.012x1023 molecules = same number of molecules How much of the 0.5M glucose solution is needed to provide 100 mg of glucose? 100 mg x 1 mole/180 grams x 1 gram/1000 mg x 1 liter/0.5 moles x 1000 ml/liter = 1.11 ml If you were to dilute with 400 ml water, what would be the ? C2=(V1 x C1)/(V2) = (100 ml x 0.5 M)/(500 ml) = 0.1 M If you were to dilute with 1.99 ml water, what would be the ? C2=(V1 x C1)/(V2) = [(10 μl x 0.5 M)/(2 ml)] x 1 ml/1000 μl = 0.0025 M = 2.5 mM How would you ? V1=(V2 xC2)/(C1)=(10mlx0.1M)/(0.5M)=2ml0.5Mglucose+8mlH2O How would you ? C1 = 0.5 M = 9% glucose [from fourth question above] V1=(V2 x C2)/(C1) = (100 ml x 1%)/(9%) = 11 ml 9% glucose + 89 ml H2O Howwouldyou mM ? C1 = 0.5 M = 500 mM glucose [from third question above] V1=(V2 x C2)/(C1) = (20 μl x 25 mM)/(500 mM) = 1 μl 500 mM glucose + 19 μl H2O How would you mM glucose/40 mM and 0.5M ? V1=(V2 x C2)/(C1) = (100 μl x 40 mM)/(500 mM) = 8 μl 500 mM glucose per 100μl final vol. V1=(V2 x C2)/(C1) = (100 μl x 40 mM)/(500 mM) = 8 μl 500 mM sucrose per 100μl final vol. Total final volume = 100μl = 8 μl 0.5 M glucose + 8 μl 0.5 M sucrose + 84 μl H2O 100 ml of the 0.5M glucose solution concentration of the diluted solution 10 μl of the 0.5M glucose solution concentration of the diluted solution prepare 10 ml of 0.1M glucose from the 0.5M glucose solution prepare 100 ml of 1% glucose from the 0.5M glucose solution prepare 20 μl of 25 glucose from the 0.5M glucose solution prepare 100 μl of 40 sucrose from the 0.5M glucose sucrose solutions
Activity 1. Prepare 2 ml of 10% solution of blue food-color dye in a test tube. (tube S [stock] = 10% dye) 10% x 2 ml = 0.2 ml dye (= 200 μl dye) + 1.8 ml H2O 2. Put 0.9 ml H2O into each of ten new test tubes. Label five of the tubes 1 through 5 and the other five A through E. 3. In tubes 1–5, prepare a two-fold serial dilution of your 10% dye solution. What is the dye concentration in each tube? #1:(0.9ml#S+0.9mlH2O)=5%; #2:(0.9ml#1+0.9mlH2O)=2.5%; #3:(0.9ml#2+0.9ml H2O) = 1.3%; #4:(0.9 ml #3 + 0.9 ml H2O) = 0.6%; #5:(0.9 ml #4 + 0.9 ml H2O) = 0.3% 4. In tubes A–E, prepare a ten-fold serial dilution of your 10% dye solution. What is the dye concentration in each of these tubes? #A:(0.1ml#S+0.9mlH2O)=1%; #B:(0.1ml#A+0.9mlH2O)=0.1%; #C:(0.1ml#B+0.9 ml H2O) = 0.01%; #D:(0.1 ml #C + 0.9 ml H2O) = 0.001%; #E:(0.1 ml #D + 0.9 ml H2O) = 0.0001% 5. Prepare a twelfth tube containing 1 ml of 0.5% dye diluted directly from the 10% stock. (tube X) (Describe how.) V1=(V2 x C2)/(C1) = (1 ml x 0.5%)/(10%) = 0.05 ml x 1000μl/ml = 50 μl 10% dye + 0.95 ml H2O 6. Arrange the twelve tubes in order of decreasing dye concentration. Does the pattern of decreasing color match your predicted calculations? #S(10%) ⇒ #1(5%) ⇒ #2(2.5%) ⇒ #3(1.3%) ⇒ #A(1%) ⇒ #4(0.6%) ⇒ #X(0.5%)⇒ #5(0.3%) ⇒ #B(0.1%) ⇒ #C(0.01%)%) ⇒ #D(0.001%)%) ⇒ #E(0.0001%) 7. Would it be more accurate to prepare a 0.1% solution by direct or serial dilution? By direct dilution, V1=(V2 x C2)/(C1) = (1 ml x 0.1%)/(10%) = 0.01 ml x 1000μl/ml = 10 μl 10% dye + 0.99 ml H2O Can more accurately measure the 100μl for serial dilutions, than 10μl for direct dilution.
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prosteve123
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Answers & Comments
How many grams of glucose would be dissolved to make 1 liter of a 0.5M glucose solution?
0.5 mole/liter x 180 grams/mole x 1 liter = 90 g
How many molecules of glucose are in that 1 liter of 0.5M glucose solution?
0.5 mole/liter x 1 liter x 6.023x1023 molecules/mole = 3.012x1023 molecules
What is the concentration of the 0.5M glucose solution expressed in mM? 0.5 mole/liter x 1000 mmoles/mole = 500 mM
What is the concentration of the 0.5M glucose solution expressed in %?
0.5 mole/liter x 180 grams/mole x 1 liter/1000ml = 0.09 g/ml = 9 g/100ml = 9%
How many grams of sucrose would be dissolved in 1 liter of a 0.5M sucrose solution? How does that compare to the grams of solute in the 0.5M glucose solution?
0.5 mole/liter x 342 grams/mole x 1 liter = 171 g = almost twice as many grams
How many molecules of sucrose in that 1 liter of 0.5M sucrose solution? How does that compare to the amount of solute in the 0.5M glucose solution?
0.5 mole/liter x 1 liter x 6.023x1023 molecules/mole = 3.012x1023 molecules = same number of molecules
How much of the 0.5M glucose solution is needed to provide 100 mg of glucose?
100 mg x 1 mole/180 grams x 1 gram/1000 mg x 1 liter/0.5 moles x 1000 ml/liter = 1.11 ml
If you were to dilute with 400 ml water, what would be the ?
C2=(V1 x C1)/(V2) = (100 ml x 0.5 M)/(500 ml) = 0.1 M
If you were to dilute with 1.99 ml water, what would be the ?
C2=(V1 x C1)/(V2) = [(10 μl x 0.5 M)/(2 ml)] x 1 ml/1000 μl = 0.0025 M = 2.5 mM
How would you ? V1=(V2 xC2)/(C1)=(10mlx0.1M)/(0.5M)=2ml0.5Mglucose+8mlH2O
How would you ? C1 = 0.5 M = 9% glucose [from fourth question above]
V1=(V2 x C2)/(C1) = (100 ml x 1%)/(9%) = 11 ml 9% glucose + 89 ml H2O
Howwouldyou mM ?
C1 = 0.5 M = 500 mM glucose [from third question above]
V1=(V2 x C2)/(C1) = (20 μl x 25 mM)/(500 mM) = 1 μl 500 mM glucose + 19 μl H2O
How would you mM glucose/40 mM and 0.5M ?
V1=(V2 x C2)/(C1) = (100 μl x 40 mM)/(500 mM) = 8 μl 500 mM glucose per 100μl final vol. V1=(V2 x C2)/(C1) = (100 μl x 40 mM)/(500 mM) = 8 μl 500 mM sucrose per 100μl final vol. Total final volume = 100μl = 8 μl 0.5 M glucose + 8 μl 0.5 M sucrose + 84 μl H2O
100 ml of the 0.5M glucose solution
concentration of the diluted solution
10 μl of the 0.5M glucose solution
concentration of the diluted solution
prepare
10 ml of 0.1M glucose
from the 0.5M glucose solution
prepare
100 ml of 1% glucose
from the 0.5M glucose solution
prepare
20 μl of 25
glucose
from the 0.5M glucose solution
prepare
100 μl of 40
sucrose
from the 0.5M glucose
sucrose solutions
Activity
1. Prepare 2 ml of 10% solution of blue food-color dye in a test tube. (tube S [stock] = 10% dye) 10% x 2 ml = 0.2 ml dye (= 200 μl dye) + 1.8 ml H2O
2. Put 0.9 ml H2O into each of ten new test tubes. Label five of the tubes 1 through 5 and the other five A through E.
3. In tubes 1–5, prepare a two-fold serial dilution of your 10% dye solution.
What is the dye concentration in each tube?
#1:(0.9ml#S+0.9mlH2O)=5%; #2:(0.9ml#1+0.9mlH2O)=2.5%; #3:(0.9ml#2+0.9ml H2O) = 1.3%; #4:(0.9 ml #3 + 0.9 ml H2O) = 0.6%; #5:(0.9 ml #4 + 0.9 ml H2O) = 0.3%
4. In tubes A–E, prepare a ten-fold serial dilution of your 10% dye solution.
What is the dye concentration in each of these tubes?
#A:(0.1ml#S+0.9mlH2O)=1%; #B:(0.1ml#A+0.9mlH2O)=0.1%; #C:(0.1ml#B+0.9 ml H2O) = 0.01%; #D:(0.1 ml #C + 0.9 ml H2O) = 0.001%; #E:(0.1 ml #D + 0.9 ml H2O) = 0.0001%
5. Prepare a twelfth tube containing 1 ml of 0.5% dye diluted directly from the 10% stock. (tube X) (Describe how.)
V1=(V2 x C2)/(C1) = (1 ml x 0.5%)/(10%) = 0.05 ml x 1000μl/ml = 50 μl 10% dye + 0.95 ml H2O
6. Arrange the twelve tubes in order of decreasing dye concentration. Does the pattern of decreasing color match your predicted calculations?
#S(10%) ⇒ #1(5%) ⇒ #2(2.5%) ⇒ #3(1.3%) ⇒ #A(1%) ⇒ #4(0.6%) ⇒ #X(0.5%)⇒ #5(0.3%) ⇒ #B(0.1%) ⇒ #C(0.01%)%) ⇒ #D(0.001%)%) ⇒ #E(0.0001%)
7. Would it be more accurate to prepare a 0.1% solution by direct or serial dilution?
By direct dilution, V1=(V2 x C2)/(C1) = (1 ml x 0.1%)/(10%) = 0.01 ml x 1000μl/ml
= 10 μl 10% dye + 0.99 ml H2O
Can more accurately measure the 100μl for serial dilutions, than 10μl for direct dilution.