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Question #0297a

1 Answer

Jan 9, 2017

$Molarity = 0.4 mol \cdot L^{- 1}$ $"Molarity"="0.4 mol"*"L"^-1$

Explanation:

$Molarity = \frac{Moles of solute}{Volume of solution}$ $"Molarity"="Moles of solute"/"Volume of solution"$

$= \frac{2 \cdot mol}{0.500 \cdot L} = 0.4 \cdot mol \cdot L^{- 1}$ $=(2*"mol")/(0.500*"L")=0.4*"mol"*"L"^-1$ .

So two questions for you:

(i) What is the concentration in $g \cdot L^{- 1}$ $"g"*"L"^-1$ ?

(ii) What are (i) the $pOH$ $"pOH"$ , and (ii) the $pH$ $"pH"$ of this solution?

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