What volume of 4.00 M $H_{2} S O_{4}$ $H_2SO_4$ is needed to prepare 500.0 mL of a solution that is 0.850 M in $H_{2} S O_{4}$ $H_2SO_4$ ?

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What volume of 4.00 M $H_{2} S O_{4}$ $H_2SO_4$ is needed to prepare 500.0 mL of a solution that is 0.850 M in $H_{2} S O_{4}$ $H_2SO_4$ ?

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Answer 1 · 2018-07-11T05:29:06

Approx. $0.10 \cdot L$ $0.10*L$ ...

Explanation:

We use the definition of concentration....

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

And thus the END molar quantity we require is...

$500 \cdot m L \times 10^{- 3} \cdot L \cdot m L^{- 1} \times 0.850 \cdot m o l \cdot L^{- 1} = 0.425 \cdot m o l$ $500*mLxx10^-3*L*mL^-1xx0.850*mol*L^-1=0.425*mol$

And since this derives from a $4.00 \cdot m o l \cdot L^{- 1}$ $4.00*mol*L^-1$ concentration of sulfuric acid...for the volume....

We take the quotient... $\frac{0.425 \cdot m o l}{4.00 \cdot m o l \cdot L^{- 1}} ≅ 0.106 \cdot L$ $(0.425*mol)/(4.00*mol*L^-1)~=0.106*L$ ...i.e. $106 \cdot m L$ $106*mL$ ...

The acid should be added to water, and never vice versa...

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What volume of 4.00 M $H_{2} S O_{4}$ $H_2SO_4$ is needed to prepare 500.0 mL of a solution that is 0.850 M in $H_{2} S O_{4}$ $H_2SO_4$ ?

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What volume of 4.00 M H2SO4H_2SO_4 is needed to prepare 500.0 mL of a solution that is 0.850 M in H2SO4H_2SO_4?

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What volume of 4.00 M $H_{2} S O_{4}$ $H_2SO_4$ is needed to prepare 500.0 mL of a solution that is 0.850 M in $H_{2} S O_{4}$ $H_2SO_4$ ?