1.(5 points) Calculate the concentration (in mM) of a solution prepared by disso
ID: 477464 • Letter: 1
Question
1.(5 points) Calculate the concentration (in mM) of a solution prepared by dissolving 3.02g of solid MgSO4 (Formula Weight: 120.37g/mole) in dH2O (final volume is 50mL)?
2.(15 points) If you dissolve 45mg glucose (Formula Weight: 180.16g/mole) in 1.0mL of water:
a.What is the concentration of glucose in M and µM?
b.If you take 5mL of this solution and dilute it to 250mL what is the concentration of the glucose in the resulting solution in µM?
c.How many nanomoles of glucose are present in 10mL of the diluted solution?
3. (5 points) Convert the concentration below to mM and µM:
a. 240mg of sucrose (Formula Weight: 342.30g/mole) per 70mL
4.(10 points) What molarity of HCl stock solution is needed so that 0.5mL of this stock solution diluted to 0.05L will yield 50mM?
5.(10 points) You have to make 100mL of 0.05M HCl from a 10M HCl stock solution, but the only measuring devices available are 100mL graduated cylinders and 10mL serological pipets. How could you accurately make the dilute HCl solution (see pgs. 8-12 in your Laboratory Manual)?
6. (25 points) You have a 0.15M stock solution of NaCl (Formula Weight: 58.4g/mole), a 0.5M stock solution of glucose (Formula Weight; 180.2g/mole), and a bottle of solid Tris base (Formula Weight: 121.1g/mole). How would prepare (be specific) 250mL of a single solution containing 50mM Tris, 25mM glucose, and 30mM NaCl.
7.(10 points) Describe how you would prepare 25mL of a 0.0250 M stock solution of potassium chloride (Formula Weight: 74.55g/mole). Using this stock solution, describe how you would prepare 100mL of the following dilutions: 2.5mM and 500M
8.(10 points) How many milligrams of MgCl2 (Formula Weight: 95.211g/mole) are contained in 250mL of 150mM MgCl2?
9.(10 points) What is the molar concentration of sodium chloride (Formula Weight: 58.44g/mole) in a 12.5% (w/v) solution?
Explanation / Answer
1.(5 points) Calculate the concentration (in mM) of a solution prepared by dissolving 3.02g of solid MgSO4 (Formula Weight: 120.37g/mole) in dH2O (final volume is 50mL)?
Answer:
The mass of MgSO4 dissolved = 3.02 grams
Moles of MgSO4 dissolved = Mass / Molar mass = 3.02 / 120.37 = 0.0251 moles = 0.0251 X 1000 mmoles
Volume of solution = 50mL = 0.05 L
Milli molarity = Millimoles / volume = 0.0251 X 1000 / 0.05 = 501.79 mM
2.(15 points) If you dissolve 45mg glucose (Formula Weight: 180.16g/mole) in 1.0mL of water:
a.What is the concentration of glucose in M and µM?
Answer:
Mass of Glucose = 45mg
Millimoles of glucose = mass in mg / molar mass = 45mg / 180.16 = 0.25 mmoles
Volume = 1mL = 10-3 L
Molarity = Moles / Volume = 0.25 X 10-3 / 10-3 = 0.25 M
Millimolarity = 250 X 103µM
b.If you take 5mL of this solution and dilute it to 250mL what is the concentration of the glucose in the resulting solution in µM?
V1 = 5mL M1 = 250 X 103 µM
V2 = 250 mL M2 = ?
We know that
M1V1 = M2V2
250 X 103µM X 5 = 250 X M2
M2 = 5000 µM
c.How many nanomoles of glucose are present in 10mL of the diluted solution?
Answer:
In 1 L (1000mL) of this diluted solution moles of glucose present = 5000 µmoles
So moles in 1 mL = 5000 / 1000 = 5 µmoles
Moles in 10mL = 50 µmoles = 50 X 103 nanomoles
3. (5 points) Convert the concentration below to mM and µM:
a. 240mg of sucrose (Formula Weight: 342.30g/mole) per 70mL
The mass of sucrose = 240mg
Molar mass = 342.30 g / mole
Moles of sucrose = mass / molar mass = 240mg / 342.30 = 0.701 mmoles
Concentration in mM will be = mmoles / volume in litres = 0.701 / 70 X 10-3 L
Concentration in mM = 10.014 mM
Concentration in µM = 10.014 X 103 µM = 10014 µM
4.(10 points) What molarity of HCl stock solution is needed so that 0.5mL of this stock solution diluted to 0.05L will yield 50mM?
Answer:
Initial concentration = ?
Initial volume = 0.5 mL
Final concentration = 50mM
Final volume = 0.05 L = 0.05 X 1000 mL
We know that
Initial concentration X initial volume = final concentration X final volume
M1 X 0.5 = 50mM X 0.05 X 1000
M1 = 5000 mM = 5 Molar
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