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Ответ:

dora8798

06.06.2020

Mathematics

Explanation:

It is known that molar mass of $V_{2}O_{5}$ is 181.88 g/mol.

We will calculate the number of vanadium atoms present in 1 mole of $V_{2}O_{5}$ as follows.

$2 \times 6.022 \times 10^{23}$ atoms

= $12.044 \times 10^{23}$ atoms

Therefore, mass of $V_{2}O_{5}$ which contains $12.044 \times 10^{23}$ atoms of vanadium = Molar mass of $V_{2}O_{5}$ = 181.88 g/mol.

Hence, we will calculate the mass of $V_{2}O_{5}$ that contains $1.0 \times 10^{6}$ vanadium atoms as follows.

$\frac{181.88 g/mol \times 1.0 \times 10^{6} atoms}{12.044 \times 10^{23}atoms}$

= $15.1 \times 10^{-17}$

Thus, we can conclude that mass of $V_{2}O_{5}$ in given situation is $15.1 \times 10^{-17}$ g/mol.

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