1. If 0.25 g of kanamycin sulfate (mw = 582.60 g/mol) is dissolved in 50 ml of w
ID: 542362 • Letter: 1
Question
1. If 0.25 g of kanamycin sulfate (mw = 582.60 g/mol) is dissolved in 50 ml of water, what is the molar concentration? What is the % (w/v) concentration?
2. Luria-Bertani (LB) Broth is a rich media commonly used to grow bacteria in culture and on agar plates. For LB agar plates, the ingredients are weighed out, added to 1 liter of water, autoclaved, allowed to cool to 55-60 degrees C, and poured into plates. The recipe to prepare 1 liter of LB agar is shown below. For each ingredient added, calculate the final concentration % (w/v) in the plates.
10 g of tryptone
5 g of yeast extract
10 g of NaCl
15 g of agar
3. If 10 ml of ampicillin (10 mg/ml) is added to the LB agar recipe shown in question 2, what is the final concentration % (w/v) of ampicillin in the plates?
What is the final molar concentration of ampicillin (mw = 371.40 g/mol) in the plates?
4. The recipe for 100 ml of a gel electrophoresis loading dye is shown below. For each ingredient, calculate the final molar concentration and the final concentration % (w/v)
0.25 g of bromophenol blue (mw = 669.96 g/mol)
0.25 g of xylene cyanol (mw = 538.60 g/mol)
50.00 g of sucrose (mw = 342 30 g/mol)
1.00 ml of 1M Tris (mw = 121.1 g/mol) (pH 8.0)
Explanation / Answer
1)
mass,m = 0.25 g
number of mol,
n = mass/molar mass
=(0.25 g)/(582.6 g/mol)
= 4.291*10^-4 mol
volume , V = 50 mL
= 5*10^-2 L
Molarity,
M = number of mol / volume in L
= 4.291*10^-4/5*10^-2
= 8.58*10^-3 M
This is molar concentration
%(w/v) = mass * 100 / volume in ml
= 0.25 g * 100 / 50 mL
= 0.50 %
Answer:
8.58*10^-3 M
0.50 %
I am allowed to answer only 1 question at a time
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.