Question

How many moles are in $23.7g$ of $KMn{O_4}$ ?

Answer 1

Potassium permanganate is a strong oxidising agent with the molecular formula of $KMn{O_4}$ It has potassium, manganese and four oxygen atoms. Thus, the molar mass of potassium permanganate is $158gm{\left( {mol} \right)^{ - 1}}$ . The number of moles of a substance can be calculated by dividing the amount of substance with the molar mass of the substance.

Formula used:
$n = \dfrac{w}{M}$
$n$ is number of moles
$w$ is weight of given substance
$M$ is molar mass of given substance

Complete answer:
Given weight of potassium permanganate is $23.7g$
Potassium permanganate is represented by the chemical formula of $KMn{O_4}$ it has one potassium atom, one manganese atom and four oxygen atoms. The molar mass of potassium is $39gm{\left( {mol} \right)^{ - 1}}$ where the atomic number of potassium is $19$ the molar mass of manganese is $55gm{\left( {mol} \right)^{ - 1}}$ whose atomic number is $24$ and the oxygen has a molar mass of $16gm{\left( {mol} \right)^{ - 1}}$ the presence of four oxygen atoms leads to the molar mass of $64gm{\left( {mol} \right)^{ - 1}}$ whose atomic number is $8$ .
Thus, the molar mass of potassium permanganate is $55 + 39 + 64 = 158gm{\left( {mol} \right)^{ - 1}}$
The given amount or weight of potassium permanganate is $23.7g$ whereas the molar mass of potassium permanganate is $158gm{\left( {mol} \right)^{ - 1}}$
Substitute the both values in the above formula
$n = \dfrac{{23.7}}{{158}}$
By the simplification, the value of number of moles will be $0.15$ moles
Thus, $0.15$ moles are present in $23.7g$ of $KMn{O_4}$.

Note:
The number of moles can be calculated from the weight of the substance and the molar mass of the substance. The weight of the substance should be present in grams only. As the molar mass of any compound will be in grams per mole.