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Benzoate
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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?