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 = $\frac{moles of solute}{volume of solution in liters}$   -(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)

$0.194 = \frac{moles of solute}{0.22}$

$moles of solute= 0.22 L\times 0.194 mole/L = 0.0427 mole$

Now, number of moles = $\frac{mass}{Molar mass}$  -(2)

Molar mass of $KMnO_4$ = $39.0983+54.938+4\times 15.99 = 158.0323 g/mol$

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

$0.0427 mole = \frac{mass}{158.0323 g/mole}$

$mass = 158.0323 g/mole\times 0.0427 mole = 6.748 g$

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