Answer
There are 16.7 g of solute in 28.4 mL of that solution.
Work Step by Step
1. Since the molarity of that solution is equal to $ 6.00 \space M \space H_3PO_4$:
$1 \space L \space solution = 6.00 \space moles \space H_3PO_4$
$\frac{1 \space L \space solution}{ 6.00 \space moles \space H_3PO_4} $ and $\frac{ 6.00 \space moles \space H_3PO_4}{1 \space L \space solution}$
2. Determine the molar mass of this compound ($H_3PO_4$), and setup the conversion factors:
Molar mass :
$H: 1.008g * 3= 3.024g $
$P: 30.97g $
$O: 16.00g * 4= 64.00g $
3.024g + 30.97g + 64.00g = 97.99g
$ \frac{1 \space mole \space (H_3PO_4)}{ 97.99 \space g \space (H_3PO_4)}$ and $ \frac{ 97.99 \space g \space (H_3PO_4)}{1 \space mole \space (H_3PO_4)}$
3. We need to convert the volume 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 $ 28.4$ mL of that solution:
$ 28.4 \space mL \space solution \times \frac{1 \space L}{1000 \space mL} \times \frac{ 6.00 \space moles \space H_3PO_4}{1 \space L \space solution} \times \frac{ 97.99 \space g \space H_3PO_4}{1 \space mole \space H_3PO_4} = 16.7 \space g \space H_3PO_4$
