Successive dilution problems

[Benzoate]

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?

 

[Bystander]

Looks good.