Question 14.10: What volume of 0.250 M solution can be prepared from 16.0 g ...

READ      Knowns      mass= 16.0 g K_2CO_3

M = 0.250 mol/L

Molar mass K_2CO_3 = 138.2 g/mol

PLAN         Find the volume that can be prepared from 16.0 g K_2CO_3
Solution map:      gK_2CO_3→ mol K_2CO_3 → L solution

CALCULATE     (16.0  \cancel{g K_2CO_3} ) ( \frac{1  \cancel{mol   K_2CO_3} }{138.2  \cancel{g   K_2CO_3} } ) ( \frac{1   L}{0.250  \cancel{mol   K_2CO_3}} ) = 0.463 L

=463 mL