Answer:
Approximately 206 grams.
Explanation:
How many moles of sulfuric acid [tex]\mathrm{H_2SO_4}[/tex] are there in this solution?
[tex]\text{Number of moles of solute} = \text{Concentration} \times \text{Volume}[/tex].
The unit for concentration "[tex]\mathrm{M}[/tex]" is equivalent to mole per liter. In other words, [tex]\rm 1\;M = 1\; mol\cdot L^{-1}[/tex]. For this solution, the concentration of [tex]\mathrm{H_2SO_4}[/tex] is [tex]\rm 0.420\;M = 0.420\; mol\cdot L^{-1}[/tex].
[tex]\begin{aligned}n(\mathrm{H_2SO_4}) &= c(\mathrm{H_2SO_4}) \cdot V\\&= \rm 0.420\;mol\cdot L^{-1}\times 5.00\; L \\&= \rm 2.10\; mol\end{aligned}[/tex].
What's the mass of that [tex]\rm 2.10\; mol[/tex] of [tex]\mathrm{H_2SO_4}[/tex]?
Start by finding the molar mass [tex]M[/tex] of [tex]\mathrm{H_2SO_4}[/tex].
Relative atomic mass data from a modern periodic table:
[tex]\displaystyle M(\mathrm{H_2SO_4}) = 2\times \underbrace{1.008}_{\mathrm{H}} + 1\times \underbrace{32.06}_{\mathrm{S}} + 4\times \underbrace{15.999}_{\mathrm{O}} = \rm 98.072\;g\cdot mol^{-1}[/tex].
[tex]\text{Mass} = \text{Quantity in moles} \times \text{Molar Mass}[/tex].
[tex]m = n \cdot M = \rm 2.10\; mol \times 98.072\;g\cdot mol^{-1} \approx 206\; g[/tex].
In other words, the chemist shall need approximately 206 grams of [tex]\mathrm{H_2SO_4}[/tex] to make this solution. As a side note, keep in mind that the 206 grams of [tex]\mathrm{H_2SO_4}[/tex] also take up considerable amount of space. Therefore it will take less than 5.00 L of water to make the 5.00 L solution.
