A chemist wishes to prepare $6.023\times {10}^{24}$molecules of Sulphur dioxide. Number of gram atoms of Sulphur required are

The chemical reaction for the preparation of sulphur dioxide from sulphur is as follows: $\mathrm{S}+{\mathrm{O}}_2\to {\mathrm{SO}}_2$

1mole ${\mathrm{SO}}_2$ molecule has  $6.023\times {10}^{23}$molecules of ${\mathrm{SO}}_2$

1 g atom of S means 1 mole of sulphur, S or $6.023\times {10}^{23}$atoms of S.

Number of gram atoms of sulphur, S needed to prepare  $6.023\times {10}^{23} {\mathrm{SO}}_2$molecules,

$\frac{6.023\times {10}^{24} {\mathrm{SO}}_2 \mathrm{molecules}}{6.023\times {10}^{23} {\mathrm{SO}}_2\mathrm{molecules}}\times 1 gram of S=10 gram atom of S$

Therefore, 10 gram atoms of S required preparing  $6.023\times {10}^{24}$molecules of ${\mathrm{SO}}_2$