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Olin [163]
3 years ago
11

Suppose you want to test the results of a transformation by growing Escherichia coli cells in LB medium containing ampicillin as

the antibiotic for selection. Ampicillin at a concentration of 100 µg/mL will kill cells that do not contain an ampicillin resistance gene, but will allow the growth of cells that have been transformed with this gene. The concentrated stock of ampicillin is 100 mg/mL. How many microliters of the ampicillin stock should you add to 50 mL of LB for a bacterial culture?
Chemistry
1 answer:
mestny [16]3 years ago
7 0

Answer:

50.0 μL

Explanation:

When a dilution is done, the mass of the solute (in this case the ampicillin) remains constant, following the Lavoiser's law that the mass is conserved. The mass is the concentration (C) multiplied by the volume (V), so if 1 is the stock solution, and 2 is the bacterial culture after the addition of the antibiotic:

m1 = m2

C1*V1 = C2*V2

C1 = 100 mg/mL = 100000 μg/mL (1 mg = 1,000μg)

C2 = 100 μg/mL

V2 = 50 mL + V1 = 50000μL + V1 (V1 in μL)

100000*V1 = 100*(50000 + V1)

1000V1 = 50000 + V1

999V1 = 50000

V1 = 50.0 μL

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The question is incomplete, here is the complete question.

A chemist prepares a solution of copper(II) fluoride by measuring out 0.0498 g of copper(II) fluoride into a 100.0mL volumetric flask and filling the flask to the mark with water.

Calculate the concentration in mol/L of the chemist's copper(II) fluoride solution. Round your answer to 3 significant digits.

<u>Answer:</u> The concentration of copper fluoride in the solution is 4.90\times 10^{-3}mol/L

<u>Explanation:</u>

To calculate the molarity of solute, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}

We are given:

Given mass of copper (II) fluoride = 0.0498 g

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Putting values in above equation, we get:

\text{Molarity of copper (II) fluoride)=\frac{0.0498\times 1000}{101.54\times 100.0}\\\\\text{Molarity of copper (II) fluoride}=4.90\times 10^{-3}mol/L

Hence, the concentration of copper fluoride in the solution is 4.90\times 10^{-3}mol/L

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