Answer
$1.15\, g$
Work Step by Step
$\text{Number of moles of KOH}=\text{Molarity}\times\text{Volume of solution in L}$
$=0.957\,M\times0.05455\,L$
$=0.0522\,mol$
1 mol $CO_{2}$ can react with 2 mol $KOH$.
$\implies \frac{0.0522\,mol}{2}=0.0261\, mol\,CO_{2}$ can react with $0.0522\, mol\, KOH$
Mass in grams of $CO_{2}$= $0.0261\,mol\times44.01\,g/mol$
$=1.15\, g$
