Solution: What mass of potassium carbonate (molar mass = 138.21 mol-1) is needed to prepare 450. mL of an aqueous solution with a potassium concentration of 0.475 M?A. 29.5 gB. 1.33 x 104 gC. 13.3 gD. 14.8 gE.

We’re being asked to calculate the mass of potassium (K₂CO₃) carbonate needed to prepare 450. mL of an aqueous solution with a potassium concentration of 0.475 M.

Recall that molarity is the ratio of the moles of solute and the volume of solution (in liters). In other words:

$\boxed{\mathbf{M}\mathbf{o}\mathbf{l}\mathbf{a}\mathbf{r}\mathbf{i}\mathbf{t}\mathbf{y}\mathbf{ }\mathbf{\left(}\mathbf{M}\mathbf{\right)}\mathbf{ }\mathbf{=}\mathbf{ }\frac{\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{e}\mathbf{s}\mathbf{ }\mathbf{o}\mathbf{f}\mathbf{ }\mathbf{s}\mathbf{o}\mathbf{l}\mathbf{u}\mathbf{t}\mathbf{e}}{\mathbf{L}\mathbf{i}\mathbf{t}\mathbf{e}\mathbf{r}\mathbf{s}\mathbf{ }\mathbf{o}\mathbf{f}\mathbf{ }\mathbf{s}\mathbf{o}\mathbf{l}\mathbf{u}\mathbf{t}\mathbf{i}\mathbf{o}\mathbf{n}}}$

We first need to determine the number of moles of potassium (K⁺).