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Questions and Answers
What is the final concentration of citrate buffer in a microtube filled with 100 μl of 0.5 M citrate buffer in a total volume of 250 μl?
How many mmol of casein are present in 5 mL of a 0.02 M casein solution?
If 15 mg of protein is diluted in 250 μl, what is its concentration in mg/ml?
What is the procedure to prepare 100 μl of a 10 mM glucose solution from a 1 M stock solution?
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What is the amount of glucose present in a 50 μl sample if the diluted serum has an absorbance of 0.7?
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What is the original concentration of a haemoglobin solution if the final concentration after several dilutions is 0.15 g/100ml?
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To prepare 200 ml of a 250 mM phosphate buffer from a 1 M stock solution, how much stock solution is needed?
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What would be the concentration of p-NPP in a microtube if 60 μl of 2.5 mM p-NPP is included in a total volume of 250 μl?
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Study Notes
Dilution Calculations
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Problem 1: To prepare 1 mL of 15 mM citrate buffer from a 100 mM stock, use the dilution formula (C1V1 = C2V2). Solve for V1 (volume of stock needed): V1 = (15 mM * 1 mL) / 100 mM = 0.15 mL. Dilute 0.15 mL of stock to a final volume of 1 mL.
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Problem 2: Making 100 µL of 10 mM glucose from a 1 M stock requires serial dilutions because pipetting volumes smaller than 2 µL is impractical. A possible approach: first dilute the stock to a more manageable concentration (e.g., 100 mM), then perform a second dilution to reach the target concentration.
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Problem 3: A solution with 15 mg protein in 250 µL has a concentration of (15 mg / 250 µL) * (1000 µL / 1 mL) = 60 mg/mL. To express this in g/L: (60 mg/mL) * (1 g / 1000 mg) * (1000 mL / 1 L) = 60 g/L.
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Problem 4: Prepare 200 mL of 250 mM phosphate buffer from a 1 M stock using C1V1 = C2V2. Solve for V1: V1 = (250 mM * 200 mL) / 1000 mM = 50 mL. Dilute 50 mL of the 1 M stock to 200 mL.
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Problem 5: To determine the amount of casein in 5 mL of 0.02 M casein, first calculate the number of moles: moles = molarity * volume = 0.02 mol/L * 0.005 L = 0.0001 mol = 0.1 mmol = 100 µmol.
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Problem 6: The final concentration is 0.15 g/100 mL or 0.0015 g/mL. Reverse the dilutions: 0.0015 g/mL * 2 * 10 * 5 = 0.15 g/mL (original concentration).
Concentration Calculations in a Mixture
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Problem 7: This problem involves calculating the final concentration of each component in a mixture. To find the final concentration, consider the total volume of the mixture (100 µL + 60 µL + 30 µL + 60 µL = 250 µL).
- 7a. Citrate buffer: The final concentration is (100 µL * 0.5 M) / 250 µL = 0.2 M.
- 7b. p-NPP: The final concentration is (60 µL * 2.5 mM) / 250 µL = 0.6 mM.
- 7c. Acid phosphatase: The final concentration is (30 µL * 0.005 mg/mL) / 250 µL = 0.0006 mg/mL.
Glucose Tolerance Test Calculations
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Problem 8: This involves calculating glucose concentration from absorbance data using a standard curve (not provided here), then accounting for dilution factors.
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8a: Determine the nmol of glucose in the 50 µL sample from the standard curve (requires the standard curve information not given).
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8b: The concentration of glucose in the diluted serum is (nmol from 8a / 50 µL) * (1000 µL/mL) = µmol/mL.
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8c: To find the concentration of glucose in the undiluted serum, multiply the diluted serum concentration by the dilution factor (x20): Concentration (µmol/mL) * 20 = Concentration (µmol/mL) in undiluted serum.
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Description
Test your understanding of dilution calculations in chemistry with this quiz. You'll solve problems involving various concentrations and volumes, using formulas like C1V1 = C2V2. Enhance your skills in preparing solutions and calculating concentrations.
