Into how many mls of water should you pour 100.0 ml of 9.00 M sulfuric acid to get a 3.00 M solution?

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Answer 1 · 2016-10-17T17:32:39

You should pour it into a $200 \cdot m L$ $200*mL$ volume of water.

I am glad you got the order of addition right. You ALWAYS add acid to water. It is well known that if you spit in acid it spits back.

Anyway, the dilution factor is simply $\frac{1}{3}$ $1/3$ .

$Concentration$ $"Concentration"$ $=$ $=$ $\frac{Moles of acid}{Volume of acid}$ $"Moles of acid"/"Volume of acid"$

$=$ $=$ $\frac{0.100 \cdot L \times 9.00 \cdot m o l \cdot L^{- 1}}{0.100 \cdot L + 0.200 \cdot L}$ $(0.100*Lxx9.00*mol*L^-1)/(0.100*L+0.200*L)$

$=$ $=$ $\frac{0.900 \cdot m o l}{0.300 \cdot L}$ $(0.900*mol)/(0.300*L)$ $=$ $=$ $3.00 \cdot m o l \cdot L^{- 1}$ $3.00*mol*L^-1$ as required.

And thus we dilute the initial $100 \cdot m L$ $100*mL$ volume of acid with a volume of $200 \cdot m L$ $200*mL$ of water.