Answer
There are 1600 g of $Al(NO_3)_3$ in 2.5 L of that solution.
Work Step by Step
1. Since the molarity of that solution is equal to $ 3.0 \space M \space Al(NO_3)_3$:
$1 \space L \space solution = 3.0 \space moles \space AlN_3O_9$
$\frac{1 \space L \space solution}{ 3.0 \space moles \space AlN_3O_9} $ and $\frac{ 3.0 \space moles \space AlN_3O_9}{1 \space L \space solution}$
2. Determine the molar mass of this compound $(Al(NO_3)_3)$, and setup the conversion factors:
Molar mass :
$Al: 26.98g $
$N: 14.01g * 3= 42.03g $
$O: 16.00g * 3 *3= 144.0g $
26.98g + 42.03g + 144.0g = 213.0g
$ \frac{1 \space mole \space (Al(NO_3)_3)}{ 213.0 \space g \space (Al(NO_3)_3)}$ and $ \frac{ 213.0 \space g \space (Al(NO_3)_3)}{1 \space mole \space (Al(NO_3)_3)}$
3. Use the conversion factors to calculate the mass of solute in $ 2.5$ L of that solution:
$ 2.5 \space L \space solution \times \frac{ 3.0 \space moles \space AlN_3O_9}{1 \space L \space solution} \times \frac{ 213.0 \space g \space AlN_3O_9}{1 \space mole \space AlN_3O_9} = 1600 \space g \space AlN_3O_9$
