Chemistry question PLEASE HELP?

How many mL of 15.0M NH3 are needed to prepare 200 mL of a 0.265M solution by dilution.

this is due in like 1.5 hours.
THanks for your help

Answer:
# of moles of NH3 in 0.265M solution=0.265/1000*200
=0.053 mol

This # of moles will be equal to the # of moles in V mL of 15.0M NH3

Therefore;
15.0/1000*V=0.053
V=3.533 mL
M x V = M x V

(15.0 M) x V = (0.265 M) x (200 mL)

V = 3.53 mL

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