Question

If $2.4088\times {10}^{26}$ electrons are there in a pure sample of $H_{2}S{O}_4$, then what is the mass of oxygen present?

• A. $512\text{g}$
• B. $812\text{g}$
• C. $1024\text{g}$
• D. $384\text{g}$

Answer 1

The correct option is A $512\text{g}$

Number of electrons in one molecule of $H_{2}S{O}_4=1\times 2+16+8\times 4=50$
So, number of electrons in 1 mole of $H_{2}S{O}_4=50\times 6.022\times {10}^{23}$
Hence, $2.4088\times {10}^{26}$ electrons = $\frac{2.4088\times {10}^{26}}{50\times 6.022\times {10}^{23}}\text{mol of}{H}_{2}S{O}_4$
= $8\text{mol of}{H}_{2}S{O}_4$
Again, one mol of $H_{2}S{O}_4$ contains 4 mol of oxygen atoms
So, 8 mol of $H_{2}S{O}_4$ will contain 32 mol of oxygen atoms
We know,
Molar mass of oxygen atom = $16\text{g/mol}$
Therefore, mass of 32 mol of oxygen atoms = $32\times 16\text{g}$
$=512\text{g}$