Answer:
Approximately [tex]53.3\; \rm g[/tex].
Explanation:
Lookup Avogadro's Number: [tex]N_{\rm A} = 6.02\times 10^{23}\; \rm mol^{-1}[/tex] (three significant figures.)
Lookup the relative atomic mass of [tex]\rm H[/tex], [tex]\rm S[/tex], and [tex]\rm O[/tex] on a modern periodic table:
- [tex]\rm H[/tex]: [tex]1.008[/tex].
- [tex]\rm S[/tex]: [tex]32.06[/tex].
- [tex]\rm O[/tex]: [tex]15.999[/tex].
(For example, the relative atomic mass of [tex]\rm H[/tex] is [tex]1.008[/tex] means that the mass of one mole of [tex]\rm H\![/tex] atoms would be approximately [tex]1.008\![/tex] grams on average.)
The question counted the number of [tex]\rm H_2SO_4[/tex] molecules without using any unit. Avogadro's Number [tex]N_{\rm A}[/tex] helps convert the unit of that count to moles.
Each mole of [tex]\rm H_2SO_4[/tex] molecules includes exactly [tex](1\; {\rm mol} \times N_\text{A}) \approx 6.02\times 10^{23}[/tex] of these [tex]\rm H_2SO_4 \![/tex] molecules.
[tex]3.27 \times 10^{23}[/tex] [tex]\rm H_2SO_4[/tex] molecules would correspond to [tex]\displaystyle n = \frac{N}{N_{\rm A}} \approx \frac{3.27 \times 10^{23}}{6.02 \times 10^{23}\; \rm mol^{-1}} \approx 0.541389\; \rm mol[/tex] of such molecules.
(Keep more significant figures than required during intermediary steps.)
The formula mass of [tex]\rm H_2SO_4[/tex] gives the mass of each mole of [tex]\rm H_2SO_4\![/tex] molecules. The value of the formula mass could be calculated using the relative atomic mass of each element:
[tex]\begin{aligned}& M({\rm H_2SO_4}) \\ &= (2 \times 1.008 + 32.06 + 4 \times 15.999)\; \rm g \cdot mol^{-1} \\ &= 98.702\; \rm g \cdot mol^{-1}\end{aligned}[/tex].
Calculate the mass of approximately [tex]0.541389\; \rm mol[/tex] of [tex]\rm H_2SO_4[/tex]:
[tex]\begin{aligned}m &= n \cdot M \\ &\approx 0.541389\; \rm mol \times 98.702\; \rm g \cdot mol^{-1}\\ &\approx 53.3\; \rm g\end{aligned}[/tex].
(Rounded to three significant figures.)
