Question

A chemist adds 220.0ml of a 0.194 m potassium permanganate kmno4 solution to a reaction flask. calculate the mass in grams of potassium permanganate the chemist has added to the flask

Answer 1

The formula of determining the molarity is:

molarity =
`(moles of solute)/(volume of solution in liters)` -(1)

Volume of solution =
`220.0 mL` (given)

Since, 1 mL = 0.001 L

Therefore, volume of the solution = 0.22 L

Molarity = 0.194 M (given)

Substituting the values in formula (1)

`0.194 = (moles of solute)/(0.22)`

`moles of solute= 0.22 L* 0.194 mole/L = 0.0427 mole`

Now, number of moles =
`(mass)/(Molar mass)` -(2)

Molar mass of
`KMnO_4` =
`39.0983+54.938+4* 15.99 = 158.0323 g/mol`

Substituting the value of number of moles and molar mass of
`KMnO_4` in equation -(2)

`0.0427 mole = (mass)/(158.0323 g/mole)`

`mass = 158.0323 g/mole* 0.0427 mole = 6.748 g`

Hence, the mass in grams of potassium permanganate the chemist has added to the flask is
`6.748 g`.