Answer
There are 4.46 g of solute in 10.5 mL of that solution.
Work Step by Step
1. Since the molarity of that solution is equal to $ 2.50 \space M \space AgNO_3$:
$1 \space L \space solution = 2.50 \space moles \space AgNO_3$
$\frac{1 \space L \space solution}{ 2.50 \space moles \space AgNO_3} $ and $\frac{ 2.50 \space moles \space AgNO_3}{1 \space L \space solution}$
2. Determine the molar mass of this compound ($AgNO_3$), and setup the conversion factors:
Molar mass :
$Ag: 107.87g $
$N: 14.01g $
$O: 16.00g * 3= 48.00g $
107.87g + 14.01g + 48.00g = 169.88g
$ \frac{1 \space mole \space (AgNO_3)}{ 169.88 \space g \space (AgNO_3)}$ and $ \frac{ 169.88 \space g \space (AgNO_3)}{1 \space mole \space (AgNO_3)}$
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 $ 10.5$ mL of that solution:
$ 10.5 \space mL \space solution \times \frac{1 \space L}{1000 \space mL} \times \frac{ 2.50 \space moles \space AgNO_3}{1 \space L \space solution} \times \frac{ 169.88 \space g \space AgNO_3}{1 \space mole \space AgNO_3} = 4.46 \space g \space AgNO_3$
