How is a $25 \cdot m L$ $25mL$ volume of a $1.50 \cdot m o l \cdot L^{- 1}$ $1.50mol*L^-1$ solution of $potassium permanganate$ $"potassium permanganate"$ prepared?

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How is a $25 \cdot m L$ $25mL$ volume of a $1.50 \cdot m o l \cdot L^{- 1}$ $1.50mol*L^-1$ solution of $potassium permanganate$ $"potassium permanganate"$ prepared?

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Answer 1 · 2017-06-02T18:26:50

Take a $5.93 \cdot g$ $5.93*g$ mass of $potassium permanganate...........$ $"potassium permanganate..........."$

Explanation:

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

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

$= \frac{\frac{5.93 \cdot g}{158.03 \cdot g \cdot m o l^{- 1}}}{25.0 \times 10^{- 3} \cdot L} = 1.50 \cdot m o l \cdot L^{- 1}$ $=((5.93*g)/(158.03*g*mol^-1))/(25.0xx10^-3*L)=1.50*mol*L^-1$ 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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How is a $25 \cdot m L$ $25mL$ volume of a $1.50 \cdot m o l \cdot L^{- 1}$ $1.50mol*L^-1$ solution of $potassium permanganate$ $"potassium permanganate"$ prepared?

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Explanation:

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How is a 25⋅mL25*mL volume of a 1.50⋅mol⋅L−11.50*mol*L^-1 solution of potassium permanganate"potassium permanganate" prepared?

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How is a $25 \cdot m L$ $25mL$ volume of a $1.50 \cdot m o l \cdot L^{- 1}$ $1.50mol*L^-1$ solution of $potassium permanganate$ $"potassium permanganate"$ prepared?