Your friend who works in a research lab on campus is overwhelmed with projects a
ID: 54883 • Letter: Y
Question
Your friend who works in a research lab on campus is overwhelmed with projects and needs your help! Suit up and make the following stock solutions (water will be your diluent).
a) If you dissolve 600 mg of glucose in 500 ul of water, how much of this solution do you need to use to make 10 ml of 22 mg/ml glucose?
b) You want to make 35 ml of a solution of potassium chloride in water at 450 mg/ml. If you have 22 ml of potassium chloride at 1.1 g/ml in water, how would you make the desired solution?
c) You have 100 ml each of 300 mg/ml sodium phosphate, 350 mg/ml sodium chloride, and 150 mg/ml potassium nitrate. You want to make one liter of a solution containing 1.0 mg/ml sodium phosphate, 13 mg/ml sodium chloride, and 15 mg/ml potassium nitrate. How much of each starting solution do you need to use and how much water do you need to add? Do you have enough materials to make this solution?
d) You have 100 ml each of 150 mg/ml human chorionic gonadotropin, 25x HCG buffer, and 450 mM potassium chloride. You want to make one liter of a solution that is 1.8 mg/ml human chorionic gonadotropin, 1x HCG buffer, and 40.5mM potassium chloride How much of each starting solution will you need to use and how much water do you need to add? Do you have enough materials to make this solution?
Explanation / Answer
Based on the given data,
a)
600 mg of dextrose in 500 micro L (500 * 10^-3 mL) water
Concentration = 800 mg/ (500 * 10^-3 mL)
= 1200 mg / mL
Total volume of solution required V1 = 10 mL
Concentration C1 = 22 mg/ mL
Original concentration C2 = 1200 mg/ mL
Volume required, V2 = C1 V1 / C2
= 22 mg / mL * 10 mL / 1200 mg/mL
= 0.183 mL
b)
Given that the concentration of stock solution is at 1.1 g/ml in water or 1100 mg/ ml. The stock solution is to be taken to prepare 35 ml of a solution of potassium chloride in water at 450 mg/ml.
Now, consider the dilution formula, C1V1 = C2V2
Where, V1 is the volume to be determined.
Substitution of values in the formula gives, (1100) (V1) = (450) (35)
V1 = 1575/110 = 14.3 ml.
Thus, 14.3 ml of stock solution must be taken and dilute upto 35ml, now the solution concentration is 450 mg/ml.
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