Successive dilution problems

  • Thread starter Benzoate
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  • #1
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Homework Statement


The following successive dilutions are applied to a stock solution that is 5.60 M sucrose

* Solution A = 46.0 mL of the stock sollution is diluted to 116 mL
* Solution B = 58.0 mL of Solution A is diluted to 248 mL
* Solution C = 87.0 mL of Solution B is diluted to 287 mL

What is the concentration of sucrose in solution C?
C(final)=C(initial)*(V(1)/V(2))*(V(3)/V(4)) * (V(5)/V(6))


Homework Equations


C(final)=C(initial)*(V(1)/V(2))*(V(3)/V(4)) * (V(5)/V(6))



The Attempt at a Solution



C(final) = (5.60 M)*(46 mL/116 mL)*(58 mL/248 mL)*(87 mL/287 mL)
C(final)= .157 M

Is that how I would calculate the final concentration?
 

Answers and Replies

  • #2
Bystander
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Looks good.
 

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