Question

How many copper atoms are in a pure copper statue with a mass of 151 kg?

Answer 1

Hint: Using Avogadro’s law, find the number of moles of iron contained in 150 kilograms and then multiply it with Avogadro’s constant to find the number of atoms.

Complete answer: In order to answer the question, we need to know about moles and molar mass. Now, matter is made up of atoms, and as matter has mass, then the atoms should have an individual mass. Molar mass of an element or compound is the mass which houses $6\times {{10}^{23}}$ particles. For, example, the hydrogen molecule has a molar mass of 2 grams. This means 2 grams of hydrogen contains $6\times {{10}^{23}}$atoms, and this number is also called the Avogadro’s number.
Number of moles of an element or a compound is the ratio of its given mass taken by the user, to its molar mass. More is the number of moles, more is the concentration of the substance. Now, let us come to the question.
Now, the molar mass of copper is $63g\,mo{{l}^{-1}}$ This means that is 63 grams of copper are taken, then it means that 1 mole of the compound is taken; that mass contains the Avogadro number of particles inside it, which houses $6\times {{10}^{23}}$ atoms. Now, we will use a simple unitary method to find out the mass which is contained in the specified number of atoms. If $6\times {{10}^{23}}$ atoms correspond to 1 mole of the compound, which is 63 grams. The number of moles of the copper atoms are:
\[
  no.\,of\,mole=\dfrac{given\,mass}{molar\,mass} \\
\Rightarrow no.\,of\,mole=\dfrac{151\times {{10}^{3}}}{63}=2376mol \\
\]
Now, we will calculate the number of copper atoms contained in 2376 moles which is:
\[no\,of\,atoms=2376\times 6\times {{10}^{23}}=1.42\times {{10}^{27}}atoms\]
Hence, 150kg of iron contains $1.42\times {{10}^{27}}$ atoms.

Note: It is to be noted that by knowing the molar mass of the elements in a compound, we can calculate the molar mass of the compound, by adding up the individual molar masses of the elements. For example, the molar mass of water(${{H}_{2}}O$) is $(2\times 1)+16=18g\,mo{{l}^{-1}}$.