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A medical lab is testing a new anticancer drug on cancer cells. The drug stock solution concentration is \(1.5 \times 10^{-9} M,\) and 1.00 \(\mathrm{mL}\) of this solution will be delivered to a dish containing \(2.0 \times 10^{5}\) cancer cells in 5.00 \(\mathrm{mL}\) of aqueous fluid. What is the ratio of drug molecules to the number of cancer cells in the dish?

Short Answer

Expert verified
The ratio of drug molecules to the number of cancer cells in the dish is approximately \(4.517 \times 10^6\).

Step by step solution

01

Calculate the number of moles of drug in the solution

To calculate the number of moles (m) of the drug in the solution, we will use the equation: m = concentration (M) * volume (L) Remember that the concentration given in the exercise is \(1.5 \times 10^{-9} M\), and the volume is 1.00 mL. Since we need the volume in liters, we will convert milliliters to liters by dividing by 1000: 1.00 mL = 0.001 L Now, we can plug in our values and calculate the number of moles: m = \(1.5 \times 10^{-9} M * 0.001 L\) m = \(1.5 \times 10^{-12}\) mol
02

Calculate the number of drug molecules in the solution

Now that we have the number of moles, we will need to convert it to drug molecules. We will use Avogadro's number (6.022 \(\times 10^{23}\)) for this conversion. To calculate the number of drug molecules (N), we will use the equation: N = moles (mol) * Avogadro's number (molecules/mol) Plugging in our values: N = \(1.5 \times 10^{-12}\) mol * \(6.022 \times 10^{23}\) molecules/mol N = \(9.033 \times 10^{11}\) drug molecules
03

Calculate the ratio of drug molecules to cancer cells

Now that we have the number of drug molecules (N) and the number of cancer cells (given in the problem as \(2.0 \times 10^5\) cells), we can calculate the ratio: Ratio = \(\frac{\text{Drug molecules}}{\text{Cancer cells}}\) Ratio = \(\frac{9.033 \times 10^{11}}{2.0 \times 10^5}\) When we divide these numbers, we get: Ratio = \(4.517 \times 10^6\) This means that there are approximately \(4.517 \times 10^6\) drug molecules for every cancer cell in the dish.

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Key Concepts

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

Avogadro's Number: The Key to Counting Molecules
Avogadro's number is a fundamental constant in chemistry. It represents the number of particles, such as atoms or molecules, within one mole of a substance. This number is approximately \(6.022 \times 10^{23}\).
Understanding Avogadro's number allows us to connect the macroscopic world we see with the microscopic world we cannot. For example, when you have a certain number of moles of a substance, Avogadro's number shows you how many individual particles are present in those moles.
In the context of our example with the anticancer drug, after determining the number of moles of the drug in the solution, Avogadro's number was used to convert moles into actual molecules. This conversion is crucial because it provides a concrete number of drug molecules involved in the experiment, which can then be compared to the number of cancer cells in the study.
The use of Avogadro's number helps in bridging the gap of scale between laboratory measurements and their application in drugs and biology.
Molar Concentration: Understanding Solutions
Molar concentration, commonly referred to as molarity, is a measure of the concentration of a solute in a solution. It is expressed in moles per liter (\(M\)).
This metric is used widely in chemistry and biology, especially when mixing substances to react or test, like the anticancer drug solution.
In the exercise, the stock solution concentration of the drug was given as \(1.5 \times 10^{-9} \text{ M}\), meaning there were \(1.5 \times 10^{-9}\) moles of drug per liter of solution.
Knowing the molarity is essential for calculating how many moles are present in a given volume of solution. Once the volume was converted into liters in the exercise, it was straightforward to find the moles of the drug present. With these moles, further calculations like determining the number of individual molecules could be performed using Avogadro's number.
Ratio Calculation: Comparing Quantities in Science
Ratios provide a way to compare two quantities, displaying how many times one number contains another. This concept can be particularly useful in scientific experiments for understanding proportions between different entities, such as molecules to cells.
In our drug experiment, once the number of drug molecules and the number of cancer cells were known, calculating their ratio highlighted how many drug molecules were available for each cancer cell.
This calculation was performed by dividing the total number of drug molecules by the total number of cancer cells in the dish:
  • Number of drug molecules = \(9.033 \times 10^{11}\)
  • Number of cancer cells = \(2.0 \times 10^5\)
Performing this division gave a ratio of approximately \(4.517 \times 10^6\), indicating that there are a vast number of drug molecules compared to each individual cancer cell.
This kind of ratio calculation is critical in biological and chemical studies to determine the adequacy or effectiveness of substances applied in treatment or testing.

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Most popular questions from this chapter

Complete and balance the following molecular equations, and then write the net ionic equation for each: \begin{equation} \begin{array}{l}{\text { (a) } \operatorname{HBr}(a q)+\mathrm{Ca}(\mathrm{OH})_{2}(a q) \longrightarrow} \\ {\text { (b) } \mathrm{Cu}(\mathrm{OH})_{2}(s)+\mathrm{HClO}_{4}(a q) \longrightarrow} \\\ {\text { (c) } \mathrm{Al}(\mathrm{OH})_{3}(s)+\mathrm{HNO}_{3}(a q) \longrightarrow}\end{array} \end{equation}

In each of the following pairs, indicate which has the higher concentration of \(\mathrm{I}^{-}\) ion: (a) 0.10 \(\mathrm{M}\) BaI \(_{2}\) or 0.25 \(\mathrm{M}\) KI solution, (b) 100 \(\mathrm{mL}\) of 0.10 \(\mathrm{M}\) KI solution or 200 \(\mathrm{mL}\) of 0.040 \(\mathrm{MZnI}_{2}\) solution, \((\mathbf{c}) 3.2 \mathrm{M}\) HI solution or a solution made by dissolving 145 g of Nal in water to make 150 \(\mathrm{mL}\) of solution.

(a) How many milliliters of a stock solution of 6.0 \(\mathrm{MHNO}_{3}\) would you have to use to prepare 110 \(\mathrm{mL}\) of 0.500 \(\mathrm{M} \mathrm{HNO}_{3} ?\) (b) If you dilute 10.0 \(\mathrm{mL}\) of the stock solution to a final volume of \(0.250 \mathrm{L},\) what will be the concentration of the diluted solution?

A sample of 7.75 g of \(\mathrm{Mg}(\mathrm{OH})_{2}\) is added to 25.0 \(\mathrm{mL}\) of 0.200 \(\mathrm{M} \mathrm{HNO}_{3}\) . (a) Write the chemical equation for the reaction that occurs. (b) Which is the limiting reactant in the reaction? (c) How many moles of \(\mathrm{Mg}(\mathrm{OH})_{2}, \mathrm{HNO}_{3},\) and \(\mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}\) are present after the reaction is complete?

(a) How many grams of ethanol, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) , should you dissolve in water to make 1.00 \(\mathrm{L}\) of vodka (which is an aqueous solution that is 6.86 \(\mathrm{M}\) ethanol)? (b) Using the density of ethanol \((0.789 \mathrm{g} / \mathrm{mL}),\) calculate the volume of ethanol you need to make 1.00 \(\mathrm{L}\) of vodka.

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