A medical lab is testing a new anticancer drug on cancer cells. The drug stock solution concentration is 1.5×10−9m, and 1.00 ml

Question

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A medical lab is testing a new anticancer drug on cancer cells. The drug stock solution concentration is 1.5×10−9m, 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?

Chemistry

2 answers:

Tems11 [2.7K] · Answer 1 · 2026-02-23T00:39:28+03:00

The concentration of the drug stock solution measures 1.5 × 10^-9 M, indicating that there are 1.5 × 10^-9 moles of the drug for every liter of solution.

To determine the number of moles in 1 ml (which is 1 × 10^-3 L), calculate: 1 × 10^-3 L × 1.5 × 10^-9 moles/1 L = 1.5 × 10^-12 moles.

Each mole of the drug consists of 6.023 × 10^23 molecules.

Thus, for 1.5 × 10^-12 moles of the drug, the corresponding number of molecules is:

1.5 × 10^-12 moles × 6.023 × 10^23 molecules/1 mole = 9.035 × 10^11 molecules.

The total number of cancer cells is 2.0 × 10^5.

The ratio hence equals the drug molecules divided by the cancer cells:

9.035 × 10^11 / 2.0 × 10^5 = 4.5 × 10^6.

alisha [2.9K] · Answer 2 · 2026-02-23T00:36:36+03:00

The ratio of drug molecules to cancer cells in the dish equals $\boxed{4.5 \times {{10}^6}}$ .

Further Explanation:

Concentration can be conveyed in various terms for different solutions. Some of these terms include:

Molarity
Mole fraction
Molality
Parts per million
Mass percent
Volume percent
Parts per billion

Molarity signifies the number of moles of solute in one liter of solution, represented as M with a unit of mol/L. The formula for molarity of a solution can be written as:

${\text{Molarity of solution}} = \dfrac{{{\text{Moles }}\left( {{\text{mol}}} \right){\text{of solute}}}}{{{\text{Volume }}\left( {\text{L}} \right){\text{ of solution}}}}$ …… (1)

Rearranging equation (1) allows for the calculation of moles of solute:

${\text{Moles of solute}} = \left( {{\text{Molarity of solution}}} \right)\left( {{\text{Volume of solution}}} \right)$ …… (2)

To find the moles of drug, replace $1.5 \times {10^{ - 9}}{\text{ M}}$ with the molarity of the solution and 1.00 mL for the solution volume in equation (2).

$\begin{aligned}{\text{Moles of drug}} &= \left( {1.5 \times {{10}^{ - 9}}{\text{ M}}} \right)\left( {1.00{\text{ mL}}} \right)\left( {\frac{{{{10}^{ - 3}}{\text{ L}}}}{{1{\text{ mL}}}}} \right) \\ &= 1.5 \times {10^{ - 12}}{\text{ mol}} \\\end{aligned}$

According to Avogadro's law, one mole of any substance contains $6.022 \times {10^{23}}{\text{ units}}$ . This can vary based on the substance, equating to molecules, atoms, or formula units. Consequently, one mole of the drug comprises $6.022 \times {10^{23}}{\text{ molecules}}$ , making it possible to compute the drug molecules in $1.5 \times {10^{ - 12}}{\text{ mol}}$ as follows:

$\begin{aligned}{\text{Molecules of drug}} &= \left( {1.5 \times {{10}^{ - 12}}{\text{ mol}}} \right)\left( {\frac{{6.022 \times {{10}^{23}}{\text{ molecules}}}}{{1{\text{ mol}}}}} \right) \\ &= 9.033 \times {10^{11}}{\text{ molecules}} \\ \end{aligned}$

The ratio of drug molecules to cancer cells in the dish can then be calculated:

$\begin{aligned}{\text{Required ratio}} &= \frac{{9.033 \times {{10}^{11}}}}{{2.0 \times {{10}^5}}} \\ &= 4.5 \times {10^6} \\ \end{aligned}$