How to create a 3 M solution of ammonia from 25% ammonia


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> We have ammonia $25\%$. We also know $\rho\_{\text{ammonia}} = \pu{0.86 \frac{g}{cm^3}}$, $\pu{25 ^\circ C}$. We want to produce $\pu{3 M}$, $\pu{0.5dm^3}$ ammonia.
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This means that there are some calculations we need to do in order to produce $\pu{3 M}$, $\pu{0.5dm^3}$ ammonia.


I try to calculate the required mass by the following method: $\text{m}=\text{C}\cdot \text{V}\cdot M \cdot 0.25$. Is this method correct?




$\text{Produce a solution of }3\;\text{M}\;0.5\;\pu{dm^3}\;\ce{NH3}$
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$\\\text{Ammonia}\;25\%\;\text{(mass-percent)}\\\rho\_{ammonia}=0.82\pu{\frac{g}{cm^3}}$


Suppose 100ml


$$0.82\times 100\times 0.25=20.5\;\pu{g}\;\ce{NH3}\text{ per 100 ml of solution}$$


$$n\_{ammonia}=\frac mM=\left(\frac{20.5}{17.031}\right)\pu{moles}$$
$$C\_{ammonia}=\frac{n\_{ammonia}}{100\times10^{-3}}\approx 12.037 \;\pu{M}.$$


To decide $V\_1$, which is needed in order to prepare the solution of $0.5 \;\pu{dm^3}$ $3\;\pu{M}$ ammonia, 


$$V\_1=\frac{C\_2V\_2}{C\_1}\approx0.1246 \pu{dm^3}=124.6\; \pu{ml}.$$


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