Problem H1: Later this semester you will be conducting a test for reducing sugar

Question

Problem H1: Later this semester you will be conducting a test for reducing sugars in solution, using a mixture called Benedict's reagent. Benedict's reagent has a sensitivity limit for detection of sugars in solution of 5 mg sugar ml. solution. Your experimental solution is in a volume of 30 ml, and the molecular weight of the sugar you will be detecting is 600. How many mmoles of the sugar will be in the experimental solution when the detection limit is reached? Problem #2: You have been hired as an orderly in an Emergency Room at a local hospital. Your first night on duty a comatose patient is wheeled in. Blood analysis readily reveals the problem: instead of the normal serum potassium concentration of 5 mM, your patient has a concentration of 1.5 mM. Since this is such an easy problem to rectify, the attending physician gives you a tube of KCI instructing you to add the necessary amount to the patient's IV drip, and hurries off to more pressing concerns. Looking at the tube in your hand, you notice the label "2M". Recalling that the combined volume of blood plus interstitial fluid in an average person is about 15 L, how much of the KCI solution should you add to bring the patient back to normal potassium levels?

Explanation / Answer

NOTE: Dear candidate you have asked four questions, but as per chegg guidelines one question one time should be solved. Still here we solve your first TWO problems.

PROBLEM 1.

Solution: Detecting limit = 5mg/mL

Volume of experimental solution = 30 mL

Hence amount of sugar that must be present in 30mL of solution to get detected = 30 X 5 = 150 mg

Therefore number of mmol of sugar = mass of sugar / Molar mass = 150 mg/600 = 0.25 mmol.

Problem 2:

Solution: Normal serum potassium concentration = 5mM

Volume of fluid = 15L

Hence normal number of moles of serum potassium = 5mM X 15L = 75mmoles

Number of moles of serum potassium of the Patient = 1.5 mM X 15 = 22.5 mmoles

Number of moles of potassium required = 75 mmoles - 22.5 mmmoles = 52.5 mmoles = 0.0525 Moles

Concentration of potassium tube to be given to patient as IV = 2M

Hence volume of this solution required = 0.0525 Moles / 2 = 0.02625 L = 26.25 mL

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