What number of magnesium ATOMS are contained in a $12 \cdot g$ $12*g$ mass of the metal?

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What number of magnesium ATOMS are contained in a $12 \cdot g$ $12*g$ mass of the metal?

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Answer 1 · 2017-03-16T16:19:56

$Number of atoms......= 3.011 \times 10^{23}$ $"Number of atoms......= "3.011xx10^23$

Explanation:

$Number of atoms of Mg$ $"Number of atoms of Mg"$ $\equiv$ $-=$ $\frac{Mass of metal}{Molar mass of metal} \times N_{A}$ $"Mass of metal"/"Molar mass of metal"xxN_A$ ,

where $N_{A} \equiv Avogadro's number = 6.022 \times 10^{23} \cdot m o l^{- 1}$ $N_A-="Avogadro's number"=6.022xx10^23*mol^-1$

And so to business,

$Number of atoms of Mg$ $"Number of atoms of Mg"$ $\equiv$ $-=$ $\frac{12 \cdot g}{24.31 \cdot g \cdot m o l^{- 1}} \times N_{A}$ $(12*g)/(24.31*g*mol^-1)xxN_A$ ,

$=1/2*cancel(mol)xx6.022xx10^23*cancel(mol^-1)$

$=3.011xx10^23$ , a $"noomber"$ as required.........

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What number of magnesium ATOMS are contained in a $12 \cdot g$ $12*g$ mass of the metal?

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What number of magnesium ATOMS are contained in a $12 \cdot g$ $12*g$ mass of the metal?