# A sample of aluminium has a mass of 6.3 g, calculate: 1. the number of mole of aluminium present....

## Question:

A sample of aluminium has a mass of {eq}6.3\ g {/eq}, calculate:

1. the number of mole of aluminium present.

2. the number of atoms of aluminium take ({eq}\rm Al = 27 {/eq}).

## 1 Mole

If {eq}6.022\times 10^{23} {/eq} particles gathered together, then we say that we have 1 mole of that particle.

If the particle gathered is atom and weighed, then the mass obtained will be Molar atomic mass.

Given

• {eq}m {/eq} = Mass of aluminium = 6.3 g
• {eq}M {/eq} = Molar mass of aluminium = 27 g/mol
• {eq}N_a {/eq} = Number of particle in 1 mole = Avogadro number = {eq}6.022\times 10^{23} {/eq}

Let

• {eq}n {/eq} = number of moles of aluminium atom in the given mass
• {eq}N {/eq} = Total number of aluminium atoms

Part (1)

We know,

{eq}\textrm{Number of moles}= \dfrac{\textrm{Given mass}}{\textrm{Molar mass}}\\ \therefore n = \dfrac{m}{M}\\ \Rightarrow n = \dfrac{6.3\ g}{27\ g/mol}\\ \Rightarrow n = 0.23\ mol {/eq}

Hence, there are 0.23 moles of AL atom in 6.3 g of it.

Part (2)

We also know,

{eq}\textrm{Total number of moles }=\textrm{Number of moles}\times \textrm{Avogadro number}\\ \therefore N = nN_a\\ \Rightarrow N = 0.23\times 6.022\times 10^{23}\\ \Rightarrow N = 1.405\times 10^{23}\ atoms {/eq}

Hence, {eq}1.405\times 10^{23} {/eq} Al atoms are there in 6.3 g of it. 