MCQ
The $pH$ of $0.1\,M\,NaOH$ is
- A$11$
- B$12$
- ✓$13$
- D$14$
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$(1)$ $2, 4-$Dinitrophenol
$(2)$ $4 -$ Nitrophenol
$(3)$ $2, 4, 5-$Trimethylphenol
$(4)$ Phenol
$(5)$ $3-$Chlorophenol
$\mathrm{Pt}(s) \mid \mathrm{H}_2(g, 1 \text { bar })\left|\mathrm{H}^{+}(a q, 1 \mathrm{M}) \| \mathrm{M}^{4+}(a q), \mathrm{M}^{2+}(a q)\right| \mathrm{Pt}(s)$
$E_{\text {cell }}=0.092 \mathrm{~V} \text { when } \frac{\left[\mathrm{M}^{2+}(a q)\right]}{\left[\mathrm{M}^{4+}(a q)\right]}=10^x$
Given : $ E_{\mathrm{M}^4 / \mathrm{M}^{2+}}^0=0.151 \mathrm{~V} ; 2.303 \frac{R T}{F}=0.059 \mathrm{~V}$
The value of $x$ is

Reason : Oxidation state of $Cl$ in $HClO_4$ is $+7$ and in $HClO_3$, it is $+ 5$.
| Column $I$ | Column $II$ |
| $(A) \,XX '$ | $(i)$ $T-$shape |
| $(B)\,XX'_3$ | $(ii)$ Pentagonal bipyramidal |
| $(C)\,XX '_5$ | $(iii)$ Linear |
| $(D)\,XX '_7$ | $(iv)$ Square pyramidal |
| $(v)$ Tetrahedral |