Question #0de15

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Question #0de15

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Answer 1 · 2016-04-15T18:41:35

$A_{r} = 58.99$ $A_r=58.99$

Explanation:

The number of moles of a substance is given by:
$\frac{m a s s (g r a m s)}{Relative atomic mass, A_{r}}$ $(mass (grams))/("Relative atomic mass", A_r)$

And one mole comprises (Avogadro's constant), $N_{A} = 6.022 \times 10^{23}$ $N_A =6.022 times 10^23$ atoms / molecules.

So number of atoms is given by:
$\frac{m a s s (g r a m s) \cdot N_{A}}{Relative atomic mass, A_{r}}$ $(mass (grams)*N_A)/("Relative atomic mass", A_r)$

In this case

$(\frac{16.26 \times 10^{- 3} g}{A_{r}}) \cdot (6.022 \times 10^{23}) = 1.66 \times 10^{20}$ $((16.26times 10^-3g)/A_r)*(6.022 times 10^23)=1.66 times 10^20$

Rearranging gives:

$A_{r} = (16.26 \times 10^{- 3} g) \cdot \frac{6.022 \times 10^{23}}{1.66 \times 10^{20}}$ $A_r= (16.26times 10^-3g)*(6.022 times 10^23)/(1.66 times 10^20)$
so
$A_{r} = 58.99$ $A_r=58.99$

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Question #0de15

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