A 0.500 L solution of 6 M HCl has to be made. How much 12 M HCI is needed?

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A 0.500 L solution of 6 M HCl has to be made. How much 12 M HCI is needed?

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Answer 1 · 2017-05-20T02:47:30

We use the relationship, $concentration = \frac{moles of solute}{volume of solution}$ $"concentration"="moles of solute"/"volume of solution"$ TWICE to get a volume of .............. $0.25 \cdot L$ $0.25*L$

Explanation:

$concentration = \frac{moles of solute}{volume of solution}$ $"concentration"="moles of solute"/"volume of solution"$

And thus $moles of solute = 6 \cdot m o l \cdot L^{- 1} \times 0.500 \cdot L = 3 \cdot m o l$ $"moles of solute"=6*mol*L^-1xx0.500*L=3*mol$ required initially.

Since this comes from a volume of $12 \cdot m o l \cdot L^{- 1}$ $12*mol*L^-1$ $H C l$ $HCl$ .....we get

$\frac{3 \cdot m o l}{12 \cdot m o l \cdot L^{- 1}} = \frac{0.25}{\frac{1}{L}} = 0.25 \cdot L$ $(3*mol)/(12*mol*L^-1)=0.25/(1/L)=0.25*L$

The concentrated hydrochloric acid used in the lab ( $36%w/w$ $"36%w/w"$ ) is $10.6 \cdot m o l \cdot L^{- 1}$ $10.6*mol*L^-1$ so this is another poor question.

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A 0.500 L solution of 6 M HCl has to be made. How much 12 M HCI is needed?

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