Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$\begin{array}{*{20}{c}}
{C{H_3}\,\,\,\,\,\,\,\,\,} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{C{H_3} - CH - C = CH}
\end{array}\xrightarrow[{{H_2}O}]{{HgS{O_4},{H_2}S{O_4}}}X$ $\xrightarrow[{(ii)\,conc.{H_2}S{O_4}/\Delta }]{{(i)\,{C_2}{H_5}MgBr,{H_2}O}}Y$
| Experiment No. | $\begin{array}{c}{[ A ]} \\ \left( mol dm ^{-3}\right)\end{array}$ | $\begin{array}{c}{[ B ]} \\ \left( mol dm ^{-3}\right)\end{array}$ | $\begin{array}{c}{[ C]} \\ \left( mol dm ^{-3}\right)\end{array}$ | Rate of reaction $\left( mol dm ^{-3} s ^{-1}\right)$ |
| $1$ | $0.2$ | $0.1$ | $0.1$ | $6.0 \times 10^{-5}$ |
| $2$ | $0.2$ | $0.2$ | $0.1$ | $6.0 \times 10^{-5}$ |
| $3$ | $0.2$ | $0.1$ | $0.2$ | $1.2 \times 10^{-4}$ |
| $4$ | $0.3$ | $0.1$ | $0.1$ | $9.0 \times 10^{-5}$ |
The rate of the reaction for $[ A ]=0.15 mol dm ^{-3},[ B ]=0.25 mol dm ^{-3}$ and $[ C ]=0.15 mol dm ^{-3}$ is found to be $Y \times 10^{-5} mol dm d ^{-3} s ^{-1}$. The value of $Y$ i. . . . . . .
