100 ml 20% (by mass) H₂SO₄ (density = 1.2 gm/ml) and 100 ml 40% (by mass) H₂SO₄ (density = 1.4 gm/ml) are mixed together. Which are the correct concentration terms for this mixture?

$\mathrm{\rho }=1.2\mathrm{g}/\mathrm{ml}$; V = 100 ml and 20% by mass is H₂SO₄
$\begin{array}{l}\mathrm{\rho }=\frac{\mathrm{m}}{\mathrm{V}}\Rightarrow \mathrm{m}=\mathrm{\rho }\times \mathrm{V}=\frac{1.2}{\mathrm{ml}}\times 100\mathrm{ml}=120\mathrm{g}\\ \therefore {\mathrm{W}}_{{\mathrm{H}}_2{\mathrm{SO}}_4}=\frac{20}{100}\times 120=24\mathrm{g}\end{array}$
Secondly; $\mathrm{\rho }=1.4\mathrm{g}/\mathrm{ml}$; 
V = 100 ml = 40% by mass
$\begin{array}{l}\Rightarrow \mathrm{\rho }=\frac{\mathrm{m}}{\mathrm{V}}\Rightarrow \mathrm{m}=1.4\times 100=140\mathrm{g}\\ \therefore {\mathrm{W}}_{{\mathrm{H}}_2{\mathrm{SO}}_4}=\frac{40}{100}\times 140=56\mathrm{g}\end{array}$
$\Rightarrow$On mixing both solutions
Final Molarity is $\frac{24}{98}+\frac{64}{98}\text{ moles in }200\mathrm{ml}$
$\therefore \mathrm{In} 1000\mathrm{ml}\Rightarrow \frac{24+56}{98}\times 5\Rightarrow \frac{80}{98}\times 5=4.08$
$\therefore$Final molarity is 4.08 M
Moles of H₂SO₄ is $\frac{24+56}{98}$
$\begin{array}{l}{\mathrm{W}}_{\text{solvent }}=(120-24)+(140-56)=180\mathrm{g}\\ \therefore \text{ in }180\mathrm{g}\text{ solvent }\Rightarrow \text{ Moles }\left({\mathrm{H}}_2{\mathrm{SO}}_4\right)=\frac{80}{98}\end{array}$
For 1000 g solvent $\frac{80}{98\times 100}\times 1000=4.54\mathrm{m}$