Answer
There are 26.6 g of $K_2CO_3$ in 0.428 L of that solution.
Work Step by Step
1. Since the molarity of that solution is equal to $ 0.450 \space M \space K_2CO_3$:
$1 \space L \space solution = 0.450 \space moles \space K_2CO_3$
$\frac{1 \space L \space solution}{ 0.450 \space moles \space K_2CO_3} $ and $\frac{ 0.450 \space moles \space K_2CO_3}{1 \space L \space solution}$
2. Determine the molar mass of this compound (K_2CO_3), and setup the conversion factors:
Molar mass :
$K: 39.10g * 2= 78.20g $
$C: 12.01g $
$O: 16.00g * 3= 48.00g $
78.20g + 12.01g + 48.00g = 138.21g
$ \frac{1 \space mole \space (K_2CO_3)}{ 138.21 \space g \space (K_2CO_3)}$ and $ \frac{ 138.21 \space g \space (K_2CO_3)}{1 \space mole \space (K_2CO_3)}$
3. Use the conversion factors to calculate the mass of solute in $ 0.428$ L of that solution:
$ 0.428 \space L \space solution \times \frac{ 0.450 \space moles \space K_2CO_3}{1 \space L \space solution} \times \frac{ 138.21 \space g \space K_2CO_3}{1 \space mole \space K_2CO_3} = 26.6 \space g \space K_2CO_3$
