If 100 ml of 1 M ${\mathrm{H}}_2{\mathrm{SO}}_4$ solution is mixed with 100 ml of 98% (w/w) ${\mathrm{H}}_2{\mathrm{SO}}_4$ solution (d = 0.1 gm/ml), hen:

$100 \mathrm{mL}\text{ of }1\mathrm{M} {\mathrm{H}}_2{\mathrm{SO}}_4+100 \mathrm{mL}(98\mathrm{\%},\mathrm{d}=0.1){\mathrm{H}}_2{\mathrm{SO}}_4$

$\begin{array}{l}=100\times 1 \mathrm{M}+100 \mathrm{mL}\times \left(\frac{98\times 10\times 0.1}{98}\right)\\ =100\times 1 \mathrm{M}+100\times 1 \mathrm{M}\end{array}$

$\therefore \left[{\mathrm{H}}_2{\mathrm{SO}}_4\right]=\frac{100\times 1+100\times 1}{200 \mathrm{mL}}=1\mathrm{M}=98\mathrm{g}/1000 \mathrm{mL}$

Mass of ${\mathrm{H}}_2{\mathrm{SO}}_4=\frac{98 \mathrm{g}\times 200 \mathrm{mL}}{1000 \mathrm{mL}}=19.6 \mathrm{g}$

Also, the concentration of  each component becomes ½ of the initial value.