How is a 25*mL volume of a 1.50*mol*L^-1 solution of "potassium permanganate" prepared?

1 Answer
anor277
Jun 2, 2017

Take a 5.93*g mass of "potassium permanganate..........."

Explanation:

Add a 25.0*mL volume of water from a volumetric pipette.......

And thus "Concentration"="Moles of potassium permanganate"/"Volume of solution"

=((5.93*g)/(158.03*g*mol^-1))/(25.0xx10^-3*L)=1.50*mol*L^-1 as required........

Note that permanganate solutions are unstable with respect to reduction, and also their deep colour often prevents the visualization of the calibration marks on your volumetric glassware.

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