100 ml 20%  (by mass) H₂SO₄ ${\mathrm{H}}_2{\mathrm{SO}}_4(\mathrm{density}=1.2\mathrm{gm}/\mathrm{ml})$ and 100 ml 40%  (by mass) ${\mathrm{H}}_2{\mathrm{SO}}_4(\mathrm{density}=1.4\mathrm{gm}/\mathrm{ml})$ are mixed together.  Which are the correct concentration terms of this Mixture?
 

$\begin{array}{l}\mathrm{\rho }=1.2\mathrm{gm}/\mathrm{mL};\mathrm{V}=1000\mathrm{mL}\text{ and }20\mathrm{\%}\text{ by mass is }{\mathrm{H}}_2{\mathrm{SO}}_4\\ \mathrm{\rho }=\frac{\mathrm{m}}{\mathrm{V}}\Rightarrow \mathrm{m}=\mathrm{\rho }\times \mathrm{V}=\frac{1.2}{\mathrm{ml}}\times 100\mathrm{ml}=120\mathrm{g}\\ {\mathrm{W}}_{{\mathrm{H}}_2{\mathrm{SO}}_4}=\frac{20}{100}\times 120=24\mathrm{g}\end{array}$

Secondly,1.4 g/mL,  v=100 mL = 40%   by mass
$$\begin{gathered} \mathrm{\rho }=\mathrm{m}/\mathrm{v}\Rightarrow \mathrm{m}=1.4\times 100=140\mathrm{g} \\ \therefore {\mathrm{w}}_{{\mathrm{H}}_2{\mathrm{SO}}_4}=\frac{40}{100}\times 140=56\mathrm{g} \end{gathered}$$
On mixing  both solution
Final molarity is $\frac{24}{98}+\frac{64}{98}$ moles in 200 mL
 $\therefore \mathrm{in}1000\mathrm{mL}=\frac{24+56}{98}\times 5=4.08$

Final molarity is 4.0 M
Moles of ${\mathrm{H}}_2{\mathrm{SO}}_4\mathrm{\hspace{0.17em}}$is $\frac{24+56}{98}$
${\mathrm{W}}_{\mathrm{solvent}}=(120-24)+(140-56)=180\mathrm{g}$
Therefore, In 180 g Solvent  $\Rightarrow \mathrm{MOLES}\left({\mathrm{H}}_2{\mathrm{SO}}_4\right)=\frac{80}{98}$ 
For 1000 g SOLVENT
$\frac{80}{98\times 180}\times 1000=4.54\mathrm{M}$