Question

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

Answer 1

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.