Question

How many liters of 1.5 M potassium permanganate could be made if 152 g of the solute are available?

Answer 1

The volume of KMnO4 produced is = 16,013.7 Litres

Step-by-step explanation:

Concentration = mass (in moles) ÷ volume (in litres)

1g = 158.03 moles

152g = 24,020.56 moles of KMnO4

1.5 M = mass (in moles) ÷ vol

⇒ Volume =
`\frac{24,020.56} {1.5}`

= 16,013.7 Litres

Answer 2

0.64 L

Step-by-step explanation:

Recall that

n= CV where n=m/M

Hence:

m/M= CV

m= given mass of solute =152g

M= molar mass of solute

C= concentration of solute in molL-1 = 1.5M

V= volume of solute =????

Molar mass of potassium permanganate= 158.034 g/mol

Thus;

152 g/158.034 gmol-1= 1.5M × V

V= 0.96/1.5

V= 0.64 L