Question

Calculate the amount 0.5 M EDTA solution (MW= 292.24 g/mol) required to make 400 mL of 70.0 mM solution?

Answer #1

To solve this answer we will have to use dilution equation which is given by:

M_{1}V_{1} = M_{2}V_{2}

where, M_{1} is initial molarity of the solution

M_{2} is final molarity of the solution

V_{1} is initial volume of the solution

V_{2} is final volume of the solution

We have been given M_{1}, M_{2} and
V_{2} in the question. Our aim is to find
V_{1}.

Substituting the values from the question in the above dilution equation we get,

0.5 V_{1} =
0.07 400 (Here units
on LHS are same as units on RHS, so you need not convert mL into
litres)

V_{1} = (0.07 400)/0.5

V_{1} = 56 mL

So to make 400 mL of 70 mM EDTA solution, you need to take 56 mL of 0.5 M EDTA solution and dilute it to 400 mL using water.

Excess (NH4)2SO4 was added to a 70.0 mL solution containing
BaCl2 (MW = 208.23 g/mol). The resulting BaSO4 (MW = 233.43 g/mol)
precipitate had a mass of 0.2790 g after it was filtered and dried.
What is the molarity of BaCl2 in the solution?

You are provided with a 0.1 mM solution of proflavine (MW:
209.25 g/mol), a 6 mM solution of DNA, and solid NaCl (MW: 58.44
g/mol). How many μL of the DNA solution would you need to make a 10
mL of the following stock solution: 1 µM proflavine, 100 mM NaCl,
and 60 µM DNA?
State your answer as an integer value (no decimal point). State
only the number, not the unit.

You are provided with a 0.1 mM solution of proflavine (MW:
209.25 g/mol), a 6 mM solution of DNA, and solid NaCl (MW: 58.44
g/mol). How many mg of NaCl would you need to make a 15 mL of the
following stock solution: 1 µM proflavine, 100 mM NaCl, and 60 µM
DNA?
Round your answer to one decimal digit. State only the number,
not the unit.

1.Show calculations for making following reagents
A. 500 ml of a 1M solution of Tris (molecular weight 121.1
g/mol)
1Pt
B. 200 ml of 0.2M EDTA (MW 372.2
g/mol)
1Pt
C. 100 ML of TE buffer (contains 10 mM Tris and 1 mm EDTA) using
A and
B

Calculate the pZn of a solution prepared by mixing 25.0 mL of
0.0100 M EDTA with 50.0 mL of 0.00500 M Zn2+. Assume that both the
Zn2+ and EDTA solutions are buffered with 0.100 M NH3 and 0.176 M
NH4Cl.

A 50.0-mL solution containing Ni2+ and Zn2+ was treated with
25.0 mL of 0.0452 M EDTA to bind all the metal. The excess
unreacted EDTA required 12.4 mL of 0.0123 M Mg2+ for complete
reaction. An excess of the reagent 2,3-dimercapto-1-propanol was
then added to displace the EDTA from zinc. Another 29.2 mL of Mg2+
were required for reaction with the liberated EDTA. Calculate the
molarity of Ni2+ and Zn2+ in the original solution.

What weight of monosodium glutamate (MW 187.1) is required to
make a 50 ml of 50 mM solution?

5.4263 g of Na2CrO4 (MW = 161.97 g/mol) is dissolved in 100.0 mL
of water. Assuming the solution has a density of 1.00 g/mL, what is
the concentration of Na (MW = 22.9898 g/mol) in the solution in
units of
a. molarity
b. parts per thousand (ppt)
c. 50.0 mL of the solution is then diluted to a final volume of
1000.0 mL. What is the concentration of Na in the diluted solution
in units of parts per million (ppm)?

A 0.6113 g mixture of KCN (MW = 65.116 g/mol) and NaCN (MW =
49.005 g/mol) was dissolved in water. AgNO3 was added to the
solution, precipitating all the CN– in solution as AgCN (MW =
133.886 g/mol). The dried precipitate weighed 1.421 g. Calculate
the weight percent of KCN and NaCN in the original sample.

A 0.6101 g mixture of KCN (MW = 65.116 g/mol) and NaCN (MW =
49.005 g/mol) was dissolved in water. AgNO3 was added to the
solution, precipitating all the CN– in solution as AgCN (MW =
133.886 g/mol). The dried precipitate weighed 1.392 g. Calculate
the weight percent of KCN and NaCN in the original sample.

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