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

Jun 2, 2017

Take a $5.93 \cdot g$ mass of $\text{potassium permanganate...........}$

#### Explanation:

Add a $25.0 \cdot m L$ volume of water from a volumetric pipette.......

And thus $\text{Concentration"="Moles of potassium permanganate"/"Volume of solution}$

$= \frac{\frac{5.93 \cdot g}{158.03 \cdot g \cdot m o {l}^{-} 1}}{25.0 \times {10}^{-} 3 \cdot L} = 1.50 \cdot m o l \cdot {L}^{-} 1$ 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.