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

1 Answer
Jun 2, 2017

Answer:

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.