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:
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 104 g
C. 13.3 g
D. 14.8 g
E. 6.65 g
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