48.0 g LICI in 424.0 mL of solution. M 15.3 g NISO4 in 424.0 mL of solution. M 19.0 g of KCN in 424.0 mL of solution. M 0.150 g of AGNO3 in 424.0 mL of solution. M
48.0 g LICI in 424.0 mL of solution. M 15.3 g NISO4 in 424.0 mL of solution. M 19.0 g of KCN in 424.0 mL of solution. M 0.150 g of AGNO3 in 424.0 mL of solution. M
Chapter13: Titrations In Analytical Chemistry
Section: Chapter Questions
Problem 13.7QAP
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Transcribed Image Text:int
48.0 g LiCI in 424.0 mL of solution.
M
15.3 g NISO4 in 424.0 mL of solution.
M
19.0 g of KCN in 424.0 mL of solution.
M
0.150 g of AgNO3 in 424.0 mL of solution.
M
WE
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A chemist wishes to make a solution using 4.21 g of CoCl3 in which the total concentration of CI ions is equal to 0.333 mol/L. What
must the final volume of the solution be?
Let's start by understanding the relationship between [CoCl3] and [CI-].
In order for [CI-] to equal 0.333 M, what must [COCI3] equal?
moles solute
Molarity is defined as
So we need to calculate the moles of CoCl3. How many moles of CoCl3 are equivalent to
L solution
4.21 g of CoCl3?
moles CoCl3
Finally we can calculate the final volume of the solution.
What final volume will produce a solution in which [CI ] = 0.333 M, using 4.21 g of CoCl3?
L solution](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff373ab9c-5892-4a94-8ba6-cbfb5b597eee%2F7537756c-ba8e-48e6-9673-d140422938a3%2Fmt06lds_processed.jpeg&w=3840&q=75)
Transcribed Image Text:edu
A chemist wishes to make a solution using 4.21 g of CoCl3 in which the total concentration of CI ions is equal to 0.333 mol/L. What
must the final volume of the solution be?
Let's start by understanding the relationship between [CoCl3] and [CI-].
In order for [CI-] to equal 0.333 M, what must [COCI3] equal?
moles solute
Molarity is defined as
So we need to calculate the moles of CoCl3. How many moles of CoCl3 are equivalent to
L solution
4.21 g of CoCl3?
moles CoCl3
Finally we can calculate the final volume of the solution.
What final volume will produce a solution in which [CI ] = 0.333 M, using 4.21 g of CoCl3?
L solution
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