Question #c66da

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Question #c66da

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Answer 1 · 2016-04-09T18:10:01

Approx. $3 \times 10^{- 3}$ $3xx10^-3$ $m o l \cdot L^{- 1}$ $mol*L^-1$

Explanation:

$Concentration, molarity$ $"Concentration, molarity"$ $=$ $=$ $\frac{Moles of solute}{Volume of solution}$ $"Moles of solute"/"Volume of solution"$ $=$ $=$ $\frac{1.56 \cdot m o l}{500 \cdot L}$ $(1.56*mol)/(500*L)$ $=$ $=$ $? ?$ $??$ $m o l \cdot L^{- 1}$ $mol*L^-1$ .

By doing the calculation this way and including the units, I am introducing a reality check on my calculation. Do the resultant units make sense? Well, they are not $L \cdot m o l^{- 1}$ $L*mol^-1$ but $m o l \cdot L^{- 1}$ $mol*L^-1$ , as is required for a concentration.

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