Procedure 5.5: Solution and Dilution work: stock soluton of 2mgow much water are
ID: 195588 • Letter: P
Question
Procedure 5.5: Solution and Dilution work: stock soluton of 2mgow much water are needed to prepare 350 ml of a 40 wg'ml solution B. A 3% fructose solution was created in 100 ml of water. For this solution calculate the number of grams, milligrams and micrograms in the solution. g/ 100 ml g/ml mg /ml /ml 3% fructose solution Using the atomic weight of fructose of 180.16 g/mol. What is the molarity (moles/L) of the above solution? C. You need to make a dilution of an over-the-counter liquid medicine according to your pet dog's weight. The medicine comes in a stock solution of 150 mg/ml. The recommended dose for a 100 pound dog is 150 mg. However, your dog only weighs 18 pounds. Calculate the correct dose for your dog (hint: use a ratio) in milligrams You would like to give your pet dog Iml total volume of medicine. To obtain the correct dose, how much medicine and how much water do you need to mix? You have micropipettes at your disposal and would like to give your pet dog the correct dose ml of 150 mg/ml medicine ml of water Covert the above to microliter volumes ul of 150 mgimi medicine ul of water Students are reminded to graph the absorbance (y-axis) vs. protein concentration (x-axis) from procedure 5. The regression line obtained will be used to complete procedure 5.3 abo the Procedure 5.2, step 1 table. Students will turn in pages 5-8 and a graph in week 6. ove and determine the protein concentration inExplanation / Answer
ANSWER A.
Required solution volume = 350 mL
Required solution concentration = 40 ug/mL
Stock solution concentration = 2 mg/mL = 2,000 ug/mL
Now, required solution contains 350 mL * 40 ug / mL = 14,000 ug of substance
So, we can take 14,000 ug / 2,000 ug/mL = 7 mL of stock solution which will contain the same amount of substance.
The remaining 350 mL - 7mL = 343 mL we can add water so as to obtain the overall required solution concentration of 40 ug/mL
ANSWER B.
3% fructose solution in 100 mL of water means 3 grams of fructose have been added.
3% fructose solution = 3 g / 100 mL (as per definition)
3% fructose solution = 0.03 g / mL (as per unitary method, 1mL will contain 1/100 grams of substance than 100 mL)
3% fructose solution = 30 mg / mL (as per conversion 1 gram = 1000 milligram)
3% fructose solution = 30,000 g / mL (as per conversion 1 milligram = 1000 microgram)
Atomic weight of fructose = 180.16 g/mol
Molarity is defined us number of moles per 1000 mL
If there are 3 g in 100 mL,
then there are 30 g in 1000 mL
These 30 grams make up 30/180.16 of a mole = 0.1665 mole
Hence the molarity is 0.1665 moles/L
ANSWER C.
Stock solution concentration = 150 mg/mL
Recommended Dose = 150 mg / 100 pounds
Dog weight = 18 pounds
Correct dose for dog = (150 / 100) * 18 = 27 mg
Required volume = 1 mL
150 mg of dose is contained within 1 mL of stock
Hence, 27 mg of dose is contained within (1/150)*27 mL of stock = 0.18 mL of stock
Remaining 1 mL - 0.18 mL = 0.82 mL can be water.
0.18 ml of 150 mg/mL medicine
0.82 mL of water
or,
180 uL of 150 mg/mL medicine
820 uL of water
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