Skip to main content
Ch.4 - Reactions in Aqueous Solution
Chapter 4, Problem 76

Calicheamicin gamma-1, C55H74IN3O21S4, is one of the most potent antibiotics known: one molecule kills one bacterial cell. Describe how you would (carefully!) prepare 25.00 mL of an aqueous calicheamicin gamma-1 solution that could kill 1.0 * 108 bacteria, starting from a 5.00 * 10-9M stock solution of the antibiotic.

Verified step by step guidance
1
Determine the number of moles of calicheamicin gamma-1 needed to kill 1.0 \times 10^8 bacteria, knowing that one molecule kills one bacterial cell. Use Avogadro's number (6.022 \times 10^{23} molecules/mol) to convert the number of molecules to moles.
Calculate the volume of the stock solution required to obtain the moles of calicheamicin gamma-1 needed. Use the formula: \( C_1V_1 = C_2V_2 \), where \( C_1 \) is the concentration of the stock solution, \( V_1 \) is the volume of the stock solution needed, \( C_2 \) is the concentration of the desired solution, and \( V_2 \) is the volume of the desired solution.
Rearrange the formula to solve for \( V_1 \): \( V_1 = \frac{C_2V_2}{C_1} \). Substitute the known values: \( C_1 = 5.00 \times 10^{-9} \text{ M} \), \( V_2 = 25.00 \text{ mL} \), and \( C_2 \) is the concentration calculated from the moles needed.
Convert the volume \( V_1 \) from liters to milliliters if necessary, as the final preparation will be in milliliters.
Carefully measure \( V_1 \) mL of the stock solution using a pipette and dilute it with distilled water in a volumetric flask to a final volume of 25.00 mL to prepare the desired solution.

Verified Solution

Video duration:
2m
This video solution was recommended by our tutors as helpful for the problem above.
Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Molarity and Dilution

Molarity (M) is a measure of concentration defined as the number of moles of solute per liter of solution. To prepare a specific concentration from a stock solution, the dilution equation (C1V1 = C2V2) is used, where C1 and V1 are the concentration and volume of the stock solution, and C2 and V2 are the desired concentration and volume of the diluted solution. Understanding this concept is crucial for accurately preparing the required solution.
Recommended video:
Guided course
00:58
Dilution Equation

Bacterial Cell Count

To determine how much calicheamicin gamma-1 is needed to kill a specific number of bacteria, one must understand the relationship between the antibiotic's potency and the bacterial cell count. In this case, the problem states that one molecule of calicheamicin can kill one bacterial cell, so to kill 1.0 * 10^8 bacteria, you need 1.0 * 10^8 molecules of the antibiotic. This translates to calculating the required molarity based on Avogadro's number.
Recommended video:
Guided course
01:21
The Electrolytic Cell

Volume Calculation

Once the required concentration of the antibiotic solution is determined, the next step is to calculate the volume of the stock solution needed to achieve this concentration in the final volume of 25.00 mL. This involves using the dilution equation to find V1, which represents the volume of the stock solution to be diluted. Accurate volume measurement is essential for ensuring the solution's effectiveness in killing the targeted bacterial cells.
Recommended video:
Guided course
02:35
Constant-Volume Calorimetry
Related Practice
Textbook Question

(a) How many milliliters of a stock solution of 6.0 M HNO3 would you have to use to prepare 110 mL of 0.500 M HNO3?

1358
views
1
rank
Textbook Question

(b) If you dilute 10.0 mL of the stock solution to a final volume of 0.250 L, what will be the concentration of the diluted solution?

1105
views
Textbook Question

medical lab is testing a new anticancer drug on cancer cells. The drug stock solution concentration is 1.5 * 10-9 M, and 1.00 mL of this solution will be delivered to a dish containing 2.0 * 105 cancer cells in 5.00 mL of aqueous fluid. What is the ratio of drug molecules to the number of cancer cells in the dish?

1409
views
Textbook Question

Pure acetic acid, known as glacial acetic acid, is a liquid with a density of 1.049 g/mL at 25 C. Calculate the molarity of a solution of acetic acid made by dissolving 20.00 mL of glacial acetic acid at 25 C in enough water to make 250.0 mL of solution.

4575
views
2
rank
Open Question
Glycerol, C3H8O3, is a substance used extensively in the manufacture of cosmetics, foodstuffs, antifreeze, and plastics. Glycerol is a water-soluble liquid with a density of 1.2656 g/mL at 15 °C. Calculate the molarity of a solution of glycerol made by dissolving 50.000 mL glycerol at 15 °C in enough water to make 250.00 mL of solution.
Open Question
You want to analyze a silver nitrate solution. What mass of KCl is needed to precipitate the silver ions from 15.0 mL of 0.200 M AgNO3 solution? (c) Given that a 0.150 M HCl(aq) solution costs $39.95 for 500 mL and that KCl costs $10/ton, which analysis procedure is more cost-effective?