44.8 liters of $H_2$  and 67.2 liters of $O_2$  are mixed at STP. The weight of  the mixture is

One mole of gas occupies a volume of 22.4 L. One mole of H₂ (gram molecular weight of H₂) holds a volume of 22.4 L. So, the mass of H₂ required to hold a volume of 44.68 L is:

$\mathrm{wt}\mathrm{of}{\mathrm{H}}_2=\frac{\mathrm{Vol}.{\mathrm{H}}_2\mathrm{at}\mathrm{STP}}{22.4}\times \mathrm{G}.\mathrm{M}.\mathrm{wt}$ 

                  $=\frac{44.8}{22.4}\times 2$

                  $=4g$

One mole of O₂ (gram molecular weight of O₂) holds a volume of 22.4 L. So, the mass of O₂ required to hold a volume of 67.2 L is:

$\mathrm{wt} \mathrm{of} {\mathrm{O}}_2=\frac{\mathrm{Vol}. \mathrm{of} {\mathrm{O}}_2 \mathrm{at} \mathrm{STP}}{22.4}\times \mathrm{G}.\mathrm{M}.\mathrm{wt}$

                 $=\frac{67.2}{22.4}\times 32$

                 $=96g$

$\therefore Totalwt=wtof{H}_2+wtof{O}_2$

                    $=4 \mathrm{g}+96 \mathrm{g}=100 \mathrm{g}$