We’re being asked to **calculate the mass of potassium (K _{2}CO_{3}) 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:

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

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

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 g

B. 1.33 x 10^{4} g

C. 13.3 g

D. 14.8 g

E. 6.65 g

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