# How many grams of manganese (IV) oxide are needed to make a 5.6 liters of a 2.1 M solution?

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The number of moles of solute required to make a solution of molarity 1 is 1 mole per liter. To make a solution of manganese dioxide with a molarity of 2.1 M will require 2.1 moles of the solute per liter of the solution.

The molar mass of manganese (IV) oxide or manganese dioxide is 86.9368 g / mole. 2.1 mole of manganese dioxide weighs 2.1*86.9368 = 182.56 g.

As we need to make 5.6 liters of the 2.1 M solution the amount of manganese dioxide required is 1022.37

Therefore we get the mass of manganese dioxide required to make 5.6 liters of a 2.1 M solution as 1022.37 g

the amount of moles = volume(ml)*molarity*10⁻³

=5600*2.1*10⁻³

=5.6*2.1

=11.76 mol

MnO₂1mol → 86.9368 g

MnO₂11.76mol → 86.9368*11.76g ~ **1022.38g**