Answer
There is a total of 6.8 g of solute in 75 mL of that solution.
Work Step by Step
1. Since the molarity of that solution is equal to $ 0.50 \space M \space C_6H_{12}O_6$:
$1 \space L \space solution = 0.50 \space mole \space C_6H_{12}O_6$
$\frac{1 \space L \space solution}{ 0.50 \space mole \space C_6H_{12}O_6} $ and $\frac{ 0.50 \space mole \space C_6H_{12}O_6}{1 \space L \space solution}$
2. Determine the molar mass of this compound ($C_6H_{12}O_6$), and setup the conversion factors:
Molar mass :
$C: 12.01g * 6= 72.06g $
$H: 1.008g * 12= 12.10g $
$O: 16.00g * 6= 96.00g $
72.06g + 12.10g + 96.00g = 180.16g
$ \frac{1 \space mole \space (C_6H_{12}O_6)}{ 180.16 \space g \space (C_6H_{12}O_6)}$ and $ \frac{ 180.16 \space g \space (C_6H_{12}O_6)}{1 \space mole \space (C_6H_{12}O_6)}$
3. 1 L = 1000 mL, therefore:
$\frac{1 \space L}{1000 \space mL}$ and $\frac{1000 \space mL}{1 \space L}$
4. Use the conversion factors to calculate the mass of solute in $ 75$ mL of that solution:
$75 \space mL \space solution \times \frac{1 \space L}{1000 \space mL} \times\frac{ 0.50 \space mole \space C_6H_{12}O_6}{1 \space L \space solution} \times \frac{ 180.16 \space g \space C_6H_{12}O_6}{1 \space mole \space C_6H_{12}O_6} = 6.8 \space g \space C_6H_{12}O_6$
