$100 \, mL$ of $0.1 \, N \, HCl$ is neutralized by $0.2 \, N \, NaOH$ solution. After adding $30 \, mL$ of $NaOH$ solution,the neutralization is continued using $0.25 \, N \, KOH$. Calculate the volume of $KOH$ in $mL$ required to complete the neutralization.
- A$16$
- B$32$
- C$35$
- D$70$
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Match the following items in List-$I$ with the corresponding results in List-$II$ (at $STP$):
List-$I$ List-$II$ (At $STP$) $(A)$ $10 \ g \ CaCO_3 \xrightarrow{\Delta} \text{decomposition}$ $(i)$ $0.224 \ L \ CO_2$ $(B)$ $1.06 \ g \ Na_2CO_3 \xrightarrow{\text{Excess } HCl} \text{reaction}$ $(ii)$ $4.48 \ L \ CO_2$ $(C)$ $2.4 \ g \ C \xrightarrow{\text{Excess } O_2} \text{combustion}$ $(iii)$ $0.448 \ L \ CO_2$ $(D)$ $0.56 \ g \ CO \xrightarrow{\text{Excess } O_2} \text{combustion}$ $(iv)$ $2.24 \ L \ CO_2$ $(v)$ $22.4 \ L \ CO_2$
| List-$I$ | List-$II$ (At $STP$) |
| $(A)$ $10 \ g \ CaCO_3 \xrightarrow{\Delta} \text{decomposition}$ | $(i)$ $0.224 \ L \ CO_2$ |
| $(B)$ $1.06 \ g \ Na_2CO_3 \xrightarrow{\text{Excess } HCl} \text{reaction}$ | $(ii)$ $4.48 \ L \ CO_2$ |
| $(C)$ $2.4 \ g \ C \xrightarrow{\text{Excess } O_2} \text{combustion}$ | $(iii)$ $0.448 \ L \ CO_2$ |
| $(D)$ $0.56 \ g \ CO \xrightarrow{\text{Excess } O_2} \text{combustion}$ | $(iv)$ $2.24 \ L \ CO_2$ |
| $(v)$ $22.4 \ L \ CO_2$ |
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