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Chapter 4: Problem 75

A medical lab is testing a new anticancer drug on cancer cells. The drug stock solution concentration is \(1.5 \times 10^{-9} \mathrm{M},\) 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 \(4.52 \times 10^{6}\) to 1.

Step by step solution

01

Calculate moles of drug in \(1.00 \mathrm{~mL}\) of stock solution

First, we need to determine the amount (in moles) of the drug present in \(1.00 \mathrm{~mL}\) of stock solution. We can use the formula: moles of drug = concentration x volume Since the stock solution concentration is \(1.5 \times 10^{-9} \mathrm{M}\) and the volume is \(1.00 \mathrm{~mL}\): moles of drug = \( (1.5 \times 10^{-9} \mathrm{M}) \times (1.00 \times 10^{-3} \mathrm{L}) \) moles of drug = \(1.5 \times 10^{-12} \mathrm{mol}\)
02

Calculate the number of drug molecules

Now that we have the moles of the drug, we can use Avogadro's number to find the number of drug molecules. Avogadro's number is \(6.022\times 10^{23}\) molecules/mol. number of drug molecules = moles of drug × Avogadro's number number of drug molecules = \( (1.5 \times 10^{-12} \mathrm{mol}) \times (6.022\times 10^{23} \mathrm{molecules/mol}) \) number of drug molecules = \(9.033\times 10^{11} \mathrm{molecules}\)
03

Calculate the ratio of drug molecules to cancer cells

Now that we have the number of drug molecules, we need to determine the ratio of drug molecules to cancer cells in the dish. There are \(2.0 \times 10^{5}\) cancer cells in the dish. Ratio of drug molecules to cancer cells = \( \frac{9.033 \times 10^{11} \mathrm{molecules}}{2.0 \times 10^{5} \mathrm{cancer~cells}}\) Ratio of drug molecules to cancer cells = \(4.517 \times 10^{6}\) The ratio of drug molecules to the number of cancer cells in the dish is approximately \(4.52 \times 10^{6}\) to 1.

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Most popular questions from this chapter

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 \(\mathrm{H}_{2} \mathrm{CO}_{3}\)

Determine the oxidation number for the indicated element in each of the following substances: (a) \(\mathrm{N}\) in \(\mathrm{N}_{2} \mathrm{H}_{4}\), (b) \(\mathrm{Nin} \mathrm{NO}_{2}\), (c) \(\mathrm{Mn}\) in \(\mathrm{MnCl}_{3}\) (d) Fe in \(\mathrm{FeSO}_{4^{\prime}}\) (e) \(\mathrm{Pt}\) in \(\mathrm{PtCl}_{4}\) (f) \(\mathrm{Cl}\) in \(\mathrm{NaClO}_{4}\).

Glycerol, \(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}_{3},\) is a substance used extensively in the manufacture of cosmetic s, foodstuffs, antifreeze, and plastics. Glycerol is a water-soluble liquid with a density of $1.2656 \mathrm{~g} / \mathrm{mL}\( at \)15^{\circ} \mathrm{C}$. Calculate the molarity of a solution of glycerol made by dissolving \(50.000 \mathrm{~mL}\) glycerol at \(15^{\circ} \mathrm{C}\) in enough water to make \(250.00 \mathrm{~mL}\) of solution.

Citric acid, \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{7}\), is a triprotic acid. It occurs naturally in citrus fruits like lemons and has applications in food flavouring and preservatives. A solution containing an unknown concentration of the acid is titrated with KOH. It requires $23.20 \mathrm{~mL}\( of \)0.500 \mathrm{M} \mathrm{KOH}$ solution to titrate all three acidic protons in \(100.00 \mathrm{~mL}\) of the citric acid solution. Write a balanced net ionic equation for the neutralization reaction, and calculate the molarity of the citric acid solution.

Write balanced molecular and net ionic equations for the reactions of (a) manganese with dilute sulfuric acid, (b) chromium with hydrobromic acid, (c) tin with hydrochloric acid, (d) aluminum with formic acid, \(\mathrm{HCOOH}\).
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