Question

Calculate the number of Aluminium ions present in 0.051g of Aluminium oxide.
( at masses al=27u,O=16u)​

Answer 1

Explanation:

Evelyn Glennie is considered one of the world's foremost percussionists and is the first and only full-time solo classical percussionist. ... Evelyn Elizabeth Ann Glennie was born July 19, 1965, the only daughter of Isobel, a school teacher, and Herbert Arthur Glennie, a beef farmer.14-Jul-2018

Answer 2

Answer:

The number of Aluminium ions present in 0.051 g of Aluminium oxide is    $6.152$ × $10^{20}$

Explanation:

The molecular mass(M) of Aluminium oxide($Al_{2}O_{3}$) = (2×27)+(3×16) = 102 g

That is 1 mole of Aluminium oxide equal to 102 g/mol

Given the mass(m) of $Al_{2}O_{3}$ = 0.051 g

we know that the number of molecules present in a compound = number of moles ×  Avogadro number

The number of moles of $Al_{2}O_{3}$  = $\frac{m}{M} = \frac{0.051}{102} =$$5.108$×$10^{-4} moles$

Therefore, number of molecules =( $5.108$ × $10^{-4}$ ) × ($6.023$ × $10^{23}$)

                                                      = $3.076$ × $10^{20} molecules$

Aluminium oxide contains two ions of Aluminium

Therefore, number of Aluminium ions = 2× $3.076$ × $10^{20}$

                                                               = $6.152$ × $10^{20}$ ions of Aluminium