MCQ
For a liquid normal boiling point is $-173\,^oC$ then at $2\ atm$ pressure it's boiling point should be nearly ......... $^oC.$ $(\Delta H_{vap} = 200\ cal/mole,\ R = 2 \ cal/mol-K,\,\,\ln \,2 = 0.7)$
- A$-73$
- B$333$
- ✓$60$
- D$103$
$\ln \frac{2}{1}=\frac{200}{2}\left[\frac{1}{100}-\frac{1}{\mathrm{T}_{2}}\right]$
$\mathrm{T}_{2}=\frac{100}{0.3}=333.33\,\mathrm{K}=60.33\,^{o} \mathrm{C}$
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| 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. . . . . . .
|
$A$ $(mol/l)$ |
$B$ $(mol/l)$ |
Rate |
| $0.05$ | $0.05$ | $1.2\times 10^{-3}$ |
| $0.10$ | $0.05$ | $2.4\times 10^{-3}$ |
| $0.05$ | $0.10$ | $1.2\times 10^{-3}$ |