To solve this problem, we need to understand the chemical reactions involved when formic acid and oxalic acid are treated with conc. $H_2SO_4. \\$1. Reaction of formic acid with conc. $H_2SO_4: \\$Formic acid $(HCOOH)$ decomposes to produce carbon monoxide $(CO)$ and water.$\\ HCOOH \rightarrow CO + H_2O \\$2. Reaction of oxalic acid with conc. $H_2SO_4: \\$Oxalic acid $(H_2C_2O_4)$ decomposes to produce carbon dioxide $(CO_2),$ carbon monoxide $(CO),$ and water.$\\ H_2C_2O_4 \rightarrow CO + CO_2 + H_2O \\$3. The gaseous mixture evolved consists of $CO$ and $CO_2. \\$When this mixture is passed through KOH pellets, $CO_2$ is absorbed by KOH, leaving only $CO. \\$Let's calculate the moles of each acid:$\\$• Molar mass of formic acid $(HCOOH) = 46 \,$g/mol$\\$• Moles of formic acid $= \frac{2.3}{46} = 0.05 \,$mol$\\$• Molar mass of oxalic acid $(H_2C_2O_4) = 90 \,$g/mol$\\$• Moles of oxalic acid $= \frac{4.5}{90} = 0.05 \,$mol$\\$4. Calculate the moles of $CO$ produced:$\\$• From formic acid: $\\$Each mole of formic acid produces 1 mole of $CO. \\$Moles of $CO$ from formic acid $= 0.05 \,$mol$\\$• From oxalic acid: $\\$Each mole of oxalic acid produces 1 mole of $CO. \\$Moles of $CO$ from oxalic acid $= 0.05 \,$mol$\\$• Total moles of $CO = 0.05 + 0.05 = 0.10 \,$mol$\\$5. Calculate the weight of $CO$ at STP:$\\$• Molar mass of $CO = 28 \,$g/mol$\\$• Weight of $CO = 0.10 \times 28 = 2.8 \,$g$\\$Therefore, the weight of the remaining product (CO) at STP is $2.8 \,$g$. \\$This corresponds to Option 1.