Calculating Final NaCl Concentration After Dilution

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The discussion centers on calculating the final concentration of NaCl after adding 2 µL of a 50 mM solution to a 30 µL reaction. The calculation involves determining the dilution factor, which is 1/16, and then applying this factor to the original concentration. The resulting concentration is calculated as 3.125 mM, which is then converted to molarity, yielding 3.125 x 10^-3 M. The response confirms that the calculations are correct.
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Hello all, I just want to confirm that I am on the right track for a certain question regarding dilution and concentration

Q: Addition of 2uL of 50mM NaCl to a 30 uL reaction gives a final NaCl concentration of ______ M.

Attempt:
so because you add 2uL to 30uL, the dilution is (2 / 2+30) = 1/16
Then I multiplied the dilution with the concentration for the new concentration
1/16 x 50mM = 3.125 mM

Finally, I converted the mM back to M
3.125 mM (1mol / 1000mmol)
answer = 3.125 x 10^-3M

does this look ok?

thank you in advance
 
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Looks correct to me.
 

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