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Calicheamicin gamma-1, \(\mathrm{C}_{55} \mathrm{H}_{74} \mathrm{IN}_{3} \mathrm{O}_{21} \mathrm{~S}_{4},\) is one of the most potent antibiotics known: one molecule kills one bacterial cell. Describe how you would (carefully!) prepare \(25.00 \mathrm{~mL}\) of an aqueous calicheamicin gamma- 1 solution that could kill \(1.0 \times 10^{8}\) bacteria, starting from a \(5.00 \times 10^{-9} \mathrm{M}\) stock solution of the antibiotic.

Short Answer

Expert verified
Use 0.332 µL of the stock solution diluted to 25.00 mL.

Step by step solution

01

Calculate number of molecules needed

Each molecule of calicheamicin gamma-1 can kill one bacterial cell. Therefore, to kill \(1.0 \times 10^8\) bacteria, we need \(1.0 \times 10^8\) molecules of calicheamicin gamma-1.
02

Calculate moles needed

Using Avogadro's number \(6.022 \times 10^{23}\) molecules/mol, calculate the moles of calicheamicin gamma-1 needed. \[\text{Moles of calicheamicin} = \frac{1.0 \times 10^8 \text{ molecules}}{6.022 \times 10^{23} \text{ molecules/mol}} = 1.66 \times 10^{-16} \text{ mol}\]
03

Use dilution formula

The dilution formula is \(C_1V_1 = C_2V_2\), 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 final solution, and \(V_2\) is the volume of the final solution. We know \(C_1 = 5.00 \times 10^{-9} \text{ M}\) and \(V_2 = 25.00 \text{ mL}\). We need to find \(V_1\).
04

Find stock solution volume needed

First calculate \(C_2\) using the moles of calicheamicin needed:\[C_2 = \frac{1.66 \times 10^{-16} \text{ mol}}{0.02500 \text{ L}} = 6.64 \times 10^{-15} \text{ M}\]Using the dilution equation \(C_1V_1 = C_2V_2\), substitute the values we know:\[5.00 \times 10^{-9} \times V_1 = 6.64 \times 10^{-15} \times 0.02500\]Solve for \(V_1\):\[V_1 = \frac{6.64 \times 10^{-15} \times 0.02500}{5.00 \times 10^{-9}} = 3.32 \times 10^{-7} \text{ L} = 0.332 \text{ \(\mu\)L}\]So, \(0.332 \text{ \(\mu\)L}\) of the stock solution is needed.
05

Conclusion and procedure

To prepare the solution, carefully measure \(0.332 \text{ \(\mu\)L}\) of the \(5.00 \times 10^{-9} \text{ M}\) stock solution using a micropipette, and add it to a container with enough water to make the total volume \(25.00 \text{ mL}\). Ensure the solution is mixed well.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Dilution Calculations
Dilution calculations are a vital part of laboratory procedures when you want to prepare solutions at specific concentrations. To find out how much of a solution you need to dilute, the formula \( C_1V_1 = C_2V_2 \) is commonly used. Here's a quick breakdown of what each symbol represents:
  • \(C_1\) is the initial concentration of your stock solution.
  • \(V_1\) is the volume of the stock solution you need to use.
  • \(C_2\) is the desired concentration of your final solution.
  • \(V_2\) is the total volume of the diluted solution you want to prepare.
To solve for any variable in this equation, you need to know the other three. By rearranging the formula, you can determine the volume you need to dilute \(V_1\). This makes it easy to find out how much of your stock solution is required to achieve the perfect concentration.
Molar Concentration
Molar concentration, often expressed as molarity (M), tells you how many moles of a substance are contained in one liter of solution. Understanding molarity is crucial because it helps you determine how potent a solution is. In the context of antibiotic preparation, this informs you how powerful the substance is at killing bacteria. For example, a solution with a molarity of \(5.00 \times 10^{-9} \text{ M}\) means there are \(5.00 \times 10^{-9}\) moles of the solute in every liter of solution. To determine the concentration for your desired solution, you can use the formula:\[M = \frac{\text{moles of solute}}{\text{liters of solution}}\]When you have both the number of moles from lab work or calculations and the total final volume in liters, you can find the molarity of the freshly prepared solution. This information is crucial when calculating dilutions.
Micropipette Usage
A micropipette is an indispensable tool in laboratory settings, especially when dealing with small volumes like micro-liters (\(\mu L\)). They are incredibly precise and enable the transfer of very small liquid volumes with accuracy and minimum error.To use a micropipette:
  • Select the appropriate micropipette for the volume you intend to measure. Pipettes are available in different sizes catering to different volume ranges.
  • Adjust the micropipette to the exact volume needed, in this case, \(0.332 \mu L\).
  • Attach a sterile tip to the micropipette. This prevents cross-contamination and ensures precision.
  • Depress the plunger to the first stop, insert the tip into the stock solution, and release the plunger slowly to draw the liquid up.
  • Place the tip into the dilution container, depress the plunger to expel the liquid, and make sure it's mixed with the solution completely.
Micropipettes are essential when accuracy is required, especially during meticulous antibiotic preparations.
Avogadro's Number
Avogadro's number, \(6.022 \times 10^{23}\) molecules/mol, is a fundamental concept in chemistry. It tells you the number of particles, usually atoms or molecules, found in one mole of substance. This constant is vital for converting between the number of molecules and the amount in moles, providing a bridge between the atomic scale and laboratory dosage.In the context of solution preparation, knowing Avogadro's number allows you to calculate how many molecules you need. For example, to kill \(1.0 \times 10^8\) bacteria, you need the same number of molecules of the antibiotic, equaling \(1.66 \times 10^{-16}\) moles when calculated with Avogadro’s number. This calculation is a crucial step in ensuring the concentration of your solution meets the necessary requirements to be effective.
Solution Preparation
Solution preparation involves several crucial steps to ensure accuracy and effectiveness. Whether preparing antibiotics or any other chemical solution, precision is key.Here's a general guideline:
  • First, calculate the required concentration and volume for your final solution using dilution calculations and molar concentration formulas.
  • Measure the precise volume of your stock solution using a micropipette, ensuring you have the correct amount needed for dilution.
  • Add the calculated volume from the stock to a mixing container.
  • Introduce the solvent (often water) to achieve the desired final volume, as planned. For example, fill to \(25.00\) mL for the solution to effectively kill bacteria as calculated.
  • Mix thoroughly to ensure the solution is homogenous so every part is uniform, increasing the solution's effectiveness.
These steps help create the perfect solution for experimental needs, maintaining safety and efficacy.

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Most popular questions from this chapter

A \(4.36-g\) sample of an unknown alkali metal hydroxide is dissolved in \(100.0 \mathrm{~mL}\) of water. An acid-base indicator is added, and the resulting solution is titrated with \(2.50 \mathrm{M} \mathrm{HCl}(a q)\) solution. The indicator changes color, signaling that the equivalence point has been reached, after \(17.0 \mathrm{~mL}\) of the hydrochloric acid solution has been added. (a) What is the molar mass of the metal hydroxide? (b) What is the identity of the alkali metal cation: \(\mathrm{Li}^{+}, \mathrm{Na}^{+}, \mathrm{K}^{+}, \mathrm{Rb}^{+},\) or \(\mathrm{Cs}^{+} ?\)

Copper exists in the form of \(\mathrm{CuFeS}_{2}\) in copper ore. Copper is isolated in a two-step process. First, \(\mathrm{CuFeS}_{2}\) is heated with \(\mathrm{SiO}_{2}\) in the presence of oxygen to form copper(I) sulfide, \(\mathrm{CuS}: 2 \mathrm{CuFeS}_{2}+2 \mathrm{SiO}_{2}(s)+4 \mathrm{O}_{2}(g) \longrightarrow \mathrm{Cu}_{2} \mathrm{~S}(s)+\) \(2 \mathrm{FeSiO}_{3}(s)+3 \mathrm{SO}_{2}(g) . \mathrm{Cu}_{2} \mathrm{~S}\) is then heated with oxygen to form copper and \(\mathrm{SO}_{2}(g) .\) (a) Write the balanced chemical equation for the second reaction. (b) Which atoms from which compounds are being oxidized, and which atoms from which compounds are being reduced? (c) How many grams of copper would be isolated from \(85.36 \mathrm{~g}\) of \(\mathrm{CuFeS}_{2}\) in copper ore?

(a) By titration, \(15.0 \mathrm{~mL}\) of \(0.1008 \mathrm{M}\) sodium hydroxide is needed to neutralize a \(0.2053-\mathrm{g}\) sample of a weak acid. What is the molar mass of the acid if it is monoprotic? (b) An elemental analysis of the acid indicates that it is composed of \(5.89 \% \mathrm{H}, 70.6 \% \mathrm{C},\) and \(23.5 \% \mathrm{O}\) by mass. What is its molecular formula?

When carbon dioxide dissolves in water, it is in equilibrium with carbonic acid \(\mathrm{H}_{2} \mathrm{CO}_{3},\) which is a weak electrolyte. What solutes are present in aqueous solution of this compound? Write the chemical equation for the ionization of \(\mathrm{H}_{2} \mathrm{CO}_{3}\)

The arsenic in a \(1.22-\mathrm{g}\) sample of a pesticide was converted to \(\mathrm{AsO}_{4}{ }^{3-}\) by suitable chemical treatment. It was then titrated using \(\mathrm{Ag}^{+}\) to form \(\mathrm{Ag}_{3} \mathrm{AsO}_{4}\) as a precipitate. (a) What is the oxidation state of \(\mathrm{As}\) in \(\mathrm{AsO}_{4}^{3-} ?(\mathbf{b})\) Name \(\mathrm{Ag}_{3} \mathrm{AsO}_{4}\) by analogy to the corresponding compound containing phosphorus in place of arsenic. (c) If it took \(25.0 \mathrm{~mL}\) of \(0.102 \mathrm{MAg}^{+}\) to reach the equivalence point in this titration, what is the mass percentage of arsenic in the pesticide?

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