What volume of 18.0 m sulfuric acid is required to prepare 19.6 L of 0.126 m H2SO4?

Answer 1

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

Answer 2

To find the volume of 18.0 M sulfuric acid needed to prepare 19.6 L of 0.126 M H2SO4, you can use the formula M1V1 = M2V2 (where M represents molarity and V represents volume). Simply plug in the values: (18.0)(V) = (0.126)(19.6) and solve for V to get the volume of 18.0 M sulfuric acid required.

Answer 3

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

Answer 4

(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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