Question

Calculate the mass of vanadium(V) oxide (V2O5) that contains a million (1.0 *10^6) vanadium atoms. Be sure your answer has a unit symbol if necessary, and round it to 3 significant digits.

Answer 1

Answer : The mass of vanadium(V) oxide will be $1.51\times 10^{-16}g$

Explanation : Given,

Number of atoms of $V_2O_5$ = $1.0\times 10^{6}$

Molar mass of $V_2O_5$ = 181.88 g/mole

In $V_2O_5$, there are 2 atoms of vanadium and 5 atoms of oxygen.

First we have to determine the moles of $V_2O_5$.

As, $2\times 6.022\times 10^{23}$ number of vanadium atom present in 1 moles of $V_2O_5$

So, $1.0\times 10^{6}$ number of vanadium atom present in $\frac{1.0\times 10^{6}}{2\times 6.022\times 10^{23}}=8.3\times 10^{-19}$ moles of $V_2O_5$

Now we have to determine the mass of $V_2O_5$.

$\text{Mass of }V_2O_5=\text{Moles of }V_2O_5\times \text{Molar mass of }V_2O_5$

$\text{Mass of }V_2O_5=(8.3\times 10^{-19}mole)\times (181.88g/mole)=1.51\times 10^{-16}g$

Therefore, the mass of vanadium(V) oxide will be $1.51\times 10^{-16}g$

Answer 2

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.