Question

The number of atoms of sulphur in $9.8$ grams of $H_{2}S{O}_4$ is:

• A. $0.6023\times {10}^{23}$
• B. $6.023\times {10}^{23}$
• C. $6\times {10}^{23}$
• D. $6.023\times {10}^{21}$

Answer 1

The correct option is A $0.6023\times {10}^{23}$

$\because 98g{H}_{2}S{O}_4$ contains$1$ mole of $H_{2}S{O}_4$.

$1$ mole of $H_{2}S{O}_4$ contain$=1$ mole of sulphur atoms.

$\therefore 9.8$ g of $H_{2}S{O}_4$ contain $=0.1$ mol of sulphur atom.

Thus, number of sulphur-atoms $=n\times N_A$ $\left(in9.8gof{H}_{2}S{O}_4\right)$

$=0.1\times 6.023\times {10}^{23}=0.6023\times {10}^{23}$.