What volume of 0.565 M $K_{2} C O_{3}$ $K_2CO_3$ contains exactly 0.115 moles $K_{2} C O_{3}$ $K_2CO_3$ ?

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Answer 1 · 2017-04-12T00:48:08

$V = 0.204$ $V=0.204$ $L$ $L$ or $20.4$ $20.4$ $m L$ $mL$

The units of molarity are $\frac{m o l}{l i t r e}$ $(mol)/(litre)$ , using this information, we can solve for the volume.

$\frac{L}{0.565 m o l} \cdot 0.115 m o l$ $L/(0.565 mol)*0.115 mol$

$L/(0.565 cancel(mol))*0.115 cancel(mol)$

$V=0.115/0.565$

$V=0.204$ $L$ or $20.4$ $mL$

What volume of 0.565 M K_2CO_3K2CO3K_2CO_3 contains exactly 0.115 moles K_2CO_3K2CO3K_2CO_3?