Answer
There are necessary 5.47 g of $HCl$ to prepare that solution.
Work Step by Step
1. Since the molarity of that solution is equal to $ 6.00 \space M \space HCl$:
$1 \space L \space solution = 6.00 \space moles \space HCl$
$\frac{1 \space L \space solution}{ 6.00 \space moles \space HCl} $ and $\frac{ 6.00 \space moles \space HCl}{1 \space L \space solution}$
2. Determine the molar mass of this compound (HCl), and setup the conversion factors:
Molar mass :
$H: 1.008g $
$Cl: 35.45g$
1.008g + 35.45g = 36.46g
$ \frac{1 \space mole \space (HCl)}{ 36.46 \space g \space (HCl)}$ and $ \frac{ 36.46 \space g \space (HCl)}{1 \space mole \space (HCl)}$
3. To convert from milliliters to liters: 1 L = 1000 mL
$\frac{1 \space L}{1000 \space mL}$ and $\frac{1000 \space mL}{1 \space L}$
4. Use the conversion factors to calculate the mass of solute in $ 25.0$ mL of that solution:
$ 25.0 \space mL \space solution \times \frac{1 \space L}{1000 \space mL} \times \frac{ 6.00 \space moles \space HCl}{1 \space L \space solution} \times \frac{ 36.46 \space g \space HCl}{1 \space mole \space HCl} = 5.47 \space g \space HCl$
