Given: Mass of $\text{H}_2\text{SO}_4$ ($w$) = $9.8 \text{ g}$ Volume of solution ($V$) = $250 \text{ cm}^3 = 250 \text{ mL} = 0.25 \text{ L}$

Molar mass of $\text{H}_2\text{SO}_4$ ($M$) = $2(1) + 32 + 4(16) = 98 \text{ g/mol}$ Since $\text{H}_2\text{SO}_4$ is a dibasic acid, its n-factor = $2$. Equivalent mass of $\text{H}_2\text{SO}_4$ ($E$) = $\frac{\text{Molar mass}}{\text{n-factor}} = \frac{98}{2} = 49 \text{ g/eq}$

Number of gram equivalents of $\text{H}_2\text{SO}_4$ = $\frac{\text{Mass}}{\text{Equivalent mass}} = \frac{9.8}{49} = 0.2 \text{ eq}$

Normality ($N$) = $\frac{\text{Number of gram equivalents of solute}}{\text{Volume of solution in L}}$ $N = \frac{0.2 \text{ eq}}{0.25 \text{ L}} = 0.8 \text{ N}$