MCQ
An alcoholic drink substance $pH = 4.7$ then $OH$ ion concentration of this solution is $({K_w} = {10^{ - 14}}\,mo{l^2}/\,{l^2})$
- A$3 \times {10^{ - 10}}$
- ✓$5 \times {10^{ - 10}}$
- C$1 \times {10^{ - 10}}$
- D$5 \times {10^{ - 8}}$
$pH + pOH = 14$;$pH = 14 - 4.7$;$pOH = 9.3$
$[O{H^ - }] = {\text{Antilog}}\,\,\,\,[ - pOH] = {\text{Antilog}}\,\,\,[ - 9.3]$
$[O{H^ - }] = 5 \times {10^{ - 10}}$
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$\mathrm{C}_6 \mathrm{H}_6(1)+\frac{15}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})+3 \mathrm{H}_2 \mathrm{O}(1) \text {. }$
The standard enthalpy of combustion of $2 \mathrm{~mol}$ of benzene is - ' $x$ ' $k J$.
$\mathrm{x}=$. . . . . . . . . .
$(1)$ standard Enthalpy of formation of $1 \mathrm{~mol}$ of $\mathrm{C}_6 \mathrm{H}_6(1)$, for the reaction $6 \mathrm{C}$ (graphite) $+3 \mathrm{H}_2(\mathrm{~g}) \rightarrow \mathrm{C}_6 \mathrm{H}_n(1)$ is $48.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$.
$(2)$ Standard Enthalpy of formation of $1 \mathrm{~mol}$ of $\mathrm{CO}_2(\mathrm{~g})$, for the reaction $\mathrm{C}$ (graphite) $+\mathrm{O}_{2(\mathrm{~g})} \rightarrow \mathrm{CO}_2(\mathrm{~g})$ is $-393.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$.
$(3)$ Standard and Enthalpy of formation of $1 \mathrm{~mol}$ of $\mathrm{H}_2 \mathrm{O}(1)$, for the reaction $\mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(1)$ is $-286 \mathrm{~kJ} \mathrm{~mol}^{-1}$.