Question

Approximately how many moles of potassium permanganate, $KMn{O}_4$, are present in $40g$ of this compound? (Atomic weights: $K=39;Mn=55;O=16$)

• A. $1mole$
• B. $0.5mole$
• C. $0.25mole$
• D. $0.1mole$

Answer 1

The correct option is C $0.25mole$

Molecular weight of $KMn{O}_4$ is [$39$$+$$55$$+$($16$$\times$$4$)], which is equal to $158grams$

$158grams$ of $KMn{O}_4$ contains $55grams$ of Potassium. (atomic weight of Potassium is $55grams$)

$40grams$ of $KMn{O}_4$ contains $\frac{55}{158}$$\ast$$40$, which is almost equal to $14grams$ of Potassium

Now,

$55grams$ of Potassium equals to $1mole$ of Potassium

$1gram$ of Potassium equals to $\frac{1}{55}mole$ of Potassium

Thus, $14grams$ of Potassium equals to $\frac{1}{55}$$\ast$$14mole$ of Potassium, which is almost equal to $0.25mole$ of Potassium

Option C is the correct answer.