Calculate the number of atoms of oxygen present in $\pu{1.3 mol}$ of $\ce{H_2SO_4}$
1 mol has $6.02\times 10^{23}$ atoms
So $\pu{1.3 mol}$ must have $1.3 \times 6.02 \times 10^{23} = 7.826 \times 10^{23}$ atoms
Since there are 4 oxygen atoms out of 7 atoms in total in $\ce{H_2SO_4}$,
${4 \over 7} \times 7.826 \times10^{23}$
So there are $4.472 \times 10^{23}$ oxygen atoms present.
However the answer key says that $3.1 \times 10^{24}$ atoms is the answer.
What went wrong?