Asked by Gabby
The capacity of a large irregularly shaped vessel cannot be computed from geometric data. As a result, it is decided that the volume be established from analytical data. The vessel is filled to its capacity mark with calcium-free water. A known small amount of a concentrated calcium solution is added, and a homogeneous solution is obtained by stirring. The volume of the solution added is insignificant with respect to volume of the vessel.
In the experiment, 350 mL of a solution 0.735 M in Ca are added to the vessel. A 50.00 mL portion of the liquid in the vessel is removed after stirring and titrated with 0.001000 M EDTA. What is the capacity of the vessel in gallons if 32.21 mL of EDTA were required to reach the endpoint? One gallon is 3.78 liters.
In the experiment, 350 mL of a solution 0.735 M in Ca are added to the vessel. A 50.00 mL portion of the liquid in the vessel is removed after stirring and titrated with 0.001000 M EDTA. What is the capacity of the vessel in gallons if 32.21 mL of EDTA were required to reach the endpoint? One gallon is 3.78 liters.
Answers
Answered by
DrBob222
mols added to vessel = M x L = 0.735 x 0.350 = 0.25725
mols titrated in 50.00 mL = M x L = 0.001 x 0.03221 = 3.221 x 10^-5 which is
3.221 x 10^-5 mols/0.05 L = 6.442 x 10^-4 mols/L.
L in vessel = 0.25725 mols x L/6.442 x 10^-4 = ?? Liters in vessel.
Convert to gallons. Check my thinking. Check my work.
volume of vessel =
mols titrated in 50.00 mL = M x L = 0.001 x 0.03221 = 3.221 x 10^-5 which is
3.221 x 10^-5 mols/0.05 L = 6.442 x 10^-4 mols/L.
L in vessel = 0.25725 mols x L/6.442 x 10^-4 = ?? Liters in vessel.
Convert to gallons. Check my thinking. Check my work.
volume of vessel =
Answered by
Anonymous
6.442 L
Answered by
Anonymous
6.442 L in mL
Answered by
Anonymous
0.00057 mL
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