How to create a 3 M solution of ammonia from 25% ammonia
$\text{Produce a solution of }3\;\text{M}\;0.5\;\pu{dm^3}\;\ce{NH3}$



$\\\text{Ammonia}\;25\%\;\text{(mass-percent)}\\\rho_{ammonia}=0.82\pu{\frac{g}{cm^3}}$

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, 

$$V_1=\frac{C_2V_2}{C_1}\approx0.1246 \pu{dm^3}=124.6\; \pu{ml}.$$