Chemistry, 10.07.2019 12:30 ameyl

Amedical 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?

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Answer from: lizr7038

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

Further Explanation:

Concentration of various solutions can be expressed by various concentration terms. Some of such concentration terms are listed below.

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

Molarity is defined as moles of solute present in one litre of solution. It is represented by M and its unit is mol/L. The expression for molarity of solution is as follows:

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

Rearrange equation (1) to calculate moles of solute.

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

Substitute $1.5 \times {10^{ - 9}}{\text{ M}}$ for molarity of solution and 1.00 mL for volume of solution in equation (2) to calculate moles of drug.

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

According to Avogadro’s law, one mole of every substance has $6.022 \times {10^{23}}{\text{ units}}$ . These units can vary from question to question and can be atoms, molecules, or formula units. Since one mole of drug also contains $6.022 \times {10^{23}}{\text{ molecules}}$ , number of molecules of drug present in $1.5 \times {10^{ - 12}}{\text{ mol}}$ can be calculated 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 the number of cancer cells in the dish can be evaluated as follows:

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

Learn more:

Calculation of volume of gas: Determine how many moles of water produce:

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Concentration

Keywords: concentration, concentration terms, solutions, molarity, Avogadro’s law, drug molecules, cancer cells, 4.5*10^6, one mole, atoms, molecules.

Answer from: Quest

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

= 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g

Answer from: Quest

1 a

2 d

3 a

4 c

explanation:

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