Volume = 33.3 mL=0.0333L ,of 3.00 m sulfuric acid that contains 9.809 g of H2SO4 solute (98.09 g/mol).
In this case, according to the given information,
it is necessary for us remember that the molarity is calculated by dividing the moles by the volume of the solution in litres so the volume is calculated by dividing moles by the given molarity.
However,
we first need to calculate the mole given the mass and molar mass of the solute, H2SO4:
n=9.809 g /98.09g/mol = 0.1 mol
Thus the volume turns out :
V=0.1mol/3.00mol/L
V=0.0333L
V=33.3mL,
Volume = 33.3 mL=0.0333L ,of 3.00 m sulfuric acid that contains 9.809 g of H2SO4 solute (98.09 g/mol).
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