How to make 0.3% ammonia water from 28-30% ammonia hydroxide

Can someone tell me how to make 0.3% ammonia water from 28-30% Ammonia hydroxide?
I am trying to make a gallon or 3.78 Liters.




Ammonium hydroxide does not exist. $\ce{NH4OH}$ is a formula that was invented in the 19th century to explain why $\ce{NH3}$ is so easily and exothermically dissolved into water, as if there was a chemical reaction with water. But it has been proven that as soon as an ion like $\ce{NH4^+}$ meets the ion $\ce{OH^-}$, the following reaction occurs : $$\ce{NH4^+ + OH- <=> NH3 + H2O}$$ But this equilibrium is strongly displaced to the right hand side. So that a solution of $\ce{NH3}$ contains only a few percent of ammonium ions.


Anyway the concentration of the so-called "concentrated $\ce{NH4OH}$" does not contain $29$% $\ce{NH4OH}$, but it contains $29$% of $\ce{NH3}$.


Now to produce a solution containing $\pu{0.3}$% $\ce{NH3}$ with this concentrated solution (density : $\pu{0.91 g/mL}$), it is necessary to dilute it $100$ times. In other words, take $\pu{37.8 g = 41.5 mL}$ into one gallon water ($\pu{3.78 L}$)


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