Question

ONLY PART B
<Ch 4 Part 1 Adaptive Follow-Up Problem 4.76 Enhanced with Feedback tem 3 Part A l Review | Constants | Periodic Table Describe how you would (caretully!) prepare 25.00 mL of an aqueous calicheamicin gamma-1 solution that could kill 1.0 x 10% bacteria, starting from a 5.00 x 10 M stock solution of the antibiotic. Complete the explanation Match the words in the left column to the appropriate blanks in the sentences on the right. Make certain each sentence is complete before submitting your answer. Words may be used more than once amicin gamma-1, CsHIN,021S. is one of th most potent antibiotics known: One molecule kills one bacterial cell You may want to reference (Pages 145 while completing this problem 148) Section 4.5 Reset Help moles The 25.00 mL of antibiotic solution needs to contain a minimum of 1.0 x 10 molecules of the drug. calculate the moles of drug this represents. The concentration of the stock solution is 5.00 x 10 Then, L stock solution mol drug/M solution: L mol/ 5.00 x 109M 1.0 x 10 mass 5.00 x 10-9 volume Previous Answers Correct The introduction states that one molecule of antibiotic kills one bacterial cell Since there are 1.0 × 10° bactenal cells, at least 1.0 × 108 molecules of antibiotic are needed. To determine the amount of stock solution we need, we must be able to compare the number of molecules of antibiotic we need to the molarity of the stock solution. The number of molecules and molarity can both be easily converted to moles for comparison; there are 6.022 x 102 molecules per mole and units of molarity are equivalent to mol/L. The molarity, or concentration, of the stock solution, is 5.00 × 10 9 M and is given in the question statement. To determine the volume of stock solution that contains 1.00 x 108 molecules of antibiotic, we must perform the calculation given at the end of the statement, which says the volume of stock solution required is equal to the number of moles required divided by the molarity of the stock solution
ONLY part B

Answer 1

Moles of drug requires = 1 times 10^8 molecule/6.022 times 10^23 molecule/mole = 1.661 times 10^-16 mole Now, use the formula L_stock = Mole/Molarity = 1.661 times 10^-16 mol /5 times 10^-9 mol/L = 3.322 times 10^-8 L = 33.22 nL