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Using V * M = constant at dilution = amount of H2SO4 [mol] in both of the solutions

V= volume [L] of the solution

M= molarity [mol/L] of the solution

So: V *18.0 = 24.9 * 0.195 gives V = ( 24.9 * 0.195 ) / 18.0 = 0.270 L

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14y ago
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13y ago

Find by multiplication how many mMol H2SO4 you will need:

0.126 (Mol/L) * 19.6 (L) = 2.470 Mol = 2470 mMol H2SO4

Divide this by the concentration H2SO4 in the concentrated acid, being: 18.0 mol/L = 18.0 mMol/mL

then you'll get:

2470 (mMol H2SO4) / 18.0 (mMol/mL) = 137.2 = 137 mL concentrated H2SO4

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11y ago

(X Volume)(18.4 M H2SO4) = (250 ml)(3.00 M H2SO4)

18.4X = 750

X = 40.8 milliliters concentrated H2SO4

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Q: What volume of 18.0 m sulfuric acid is required to prepare 19.6 L of 0.126 m H2SO4?
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0.1 N means you have a hydronium ion concentration of 0.1 mole/L. Since sulfuric acid is a diprotic acid, you get 2 moles of ionizable protons per mole a sulfuric acid. Meaning, you only need a concentration of 0.05 M sulfuric acid to obtain 0.1 N sulfuric acid. [10 L] * [0.1 mole H+/L] * [1 mole H2SO4/2 mole H+] * [98.08 g/mole H2SO4] * [1 mL/1.84 g H2SO4] = 26.65 mL H2SO4 required to make 10 L 0.1 N H2SO4.


How many moles of sulfuric acid are in 500 mL of 0.324 M sulfuric acid solution?

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