First, we need to find the molarity of the concentrated $\text{H}_2\text{SO}_4$ solution. Consider $1 \text{ L}$ ($1000 \text{ mL}$) of this solution. Using the given density ($1.80 \text{ g mL}^{-1}$), the mass of $1 \text{ L}$ of solution is: $\text{Mass} = \text{Volume} \times \text{Density} = 1000 \text{ mL} \times 1.80 \text{ g mL}^{-1} = 1800 \text{ g}$ . Since the solution is $98\% \text{ H}_2\text{SO}_4$ by mass, the mass of $\text{H}_2\text{SO}_4$ in $1 \text{ L}$ of solution is: $\text{Mass of H}_2\text{SO}_4 = \frac{98}{100} \times 1800 \text{ g} = 1764 \text{ g}$ . The molar mass of $\text{H}_2\text{SO}_4$ is $2(1) + 32 + 4(16) = 98 \text{ g mol}^{-1}$ . Number of moles of $\text{H}_2\text{SO}_4$ in $1 \text{ L}$ of solution $= \frac{1764 \text{ g}}{98 \text{ g mol}^{-1}} = 18 \text{ mol}$. Therefore, the molarity ($M_1$) of the concentrated acid is $18 \text{ M}$ .

To find the volume ($V_1$) of this acid required to make $1 \text{ L}$ ($1000 \text{ mL}$) of $0.1 \text{ M}$ ($M_2$) solution, we use the dilution formula: $M_1V_1 = M_2V_2$ $18 \text{ M} \times V_1 = 0.1 \text{ M} \times 1000 \text{ mL}$ $V_1 = \frac{100}{18} \text{ mL} = 5.55 \text{ mL}$.