229098
Degree of dissociation of an acid $\mathrm{HCl}$ is $95 \%$. $0.192 \mathrm{~g}$ of the acid is present in $0.5 \mathrm{~L}$ of solution. The $\mathrm{pH}$ of the solution is :
229099
For the following equilibrium $\mathrm{N}_{2} \mathrm{O}_{4} \rightleftharpoons \quad 2 \mathrm{NO}_{2}$ in gaseous phase, $\mathrm{NO}_{2}$ is $50 \%$ of the total volume, when equilibrium is set up. Hence, percent dissociation of $\mathrm{N}_{2} \mathrm{O}_{4}$ is :
1 $50 \%$
2 $25 \%$
3 $66.66 \%$
4 $33.33 \%$
Explanation:
The dissociation of $\mathrm{N}_{2} \mathrm{O}_{4}$ is- $\mathrm{N}_{2} \mathrm{O}_{4} \rightleftharpoons \quad 2 \mathrm{NO}_{2}$ Initial $\quad 1 \quad 0$ At equilibrium $1-\mathrm{x} \quad 2 \mathrm{x}$ Total numbers of moles $=1-x+2 x=1+x$ $\therefore \%$ of $\mathrm{NO}_{2}$ by Volume $=\frac{2 \mathrm{x}}{1+\mathrm{x}} \times 100=50$ $\mathrm{x}=\frac{1}{3}$ Hence, $\%$ dissociation $=33.33 \%$
BCECE-2017
Chemical Equilibrium
229101
For which among the following equimolar aqueous solutions Van't Hoff factor has the lowest value?
1 Aluminium chloride
2 Potassium sulphate
3 Ammonium chloride
4 Urea
Explanation:
For the aqueous solution of urea, Van't Hoff factor has the lowest value. Urea is non-electrolyte and does not dissociate in aqueous solution. Hence, Vant Hoff factor is 1. For Aluminium chloride $\mathrm{AlCl}_{3-}$ $\begin{gathered} \mathrm{AlCl}_{3} \rightarrow \mathrm{Al}^{3+}+3 \mathrm{Cl}^{-} \\ i=4 \end{gathered}$ For potassium sulphate $\mathrm{K}_{2} \mathrm{SO}_{4}$ $\begin{gathered} \mathrm{K}_{2} \mathrm{SO}_{4} \rightarrow 2 \mathrm{~K}^{+}+\mathrm{SO}_{4}^{2-} \\ i=3 \end{gathered}$
MHT CET-2017
Chemical Equilibrium
229103
When $\mathrm{PCl}_{5}$ is heated it gasifies and dissociates into $\mathrm{PCl}_{3}$ and $\mathrm{Cl}_{2}$. The density of the gas mixture at $200^{\circ} \mathrm{C}$ is 70.2 . What is the degree of dissociation of $\mathrm{PCl}_{5}$ at $200^{\circ} \mathrm{C}$ ?
1 0.485
2 0.242
3 0.845
4 0.542
Explanation:
The dissociation of $\mathrm{PCl}_{5}$ is$\mathrm{PCl}_{5}(\mathrm{~g}) \rightarrow \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})$ The initial vapour density will be same, it would be $\mathrm{M}_{\mathrm{PCl}_{5}} / 2$. $\therefore$ Initial vapour density $=(31+5 \times 35.5) / 2=104.25$ Vapour density at equilibrium at $200^{\circ} \mathrm{C}=70.2$ $\therefore$ Total moles at equilibrium $\begin{aligned} & (1+\alpha)=\frac{\text { Vapour density initial }}{\text { Vapour density at equilibrium }} \\ & (1+\alpha)=\frac{104.25}{70.2}=1.485 \\ & 1+\alpha=1.485 \\ & \alpha=0.485 \end{aligned}$
229098
Degree of dissociation of an acid $\mathrm{HCl}$ is $95 \%$. $0.192 \mathrm{~g}$ of the acid is present in $0.5 \mathrm{~L}$ of solution. The $\mathrm{pH}$ of the solution is :
229099
For the following equilibrium $\mathrm{N}_{2} \mathrm{O}_{4} \rightleftharpoons \quad 2 \mathrm{NO}_{2}$ in gaseous phase, $\mathrm{NO}_{2}$ is $50 \%$ of the total volume, when equilibrium is set up. Hence, percent dissociation of $\mathrm{N}_{2} \mathrm{O}_{4}$ is :
1 $50 \%$
2 $25 \%$
3 $66.66 \%$
4 $33.33 \%$
Explanation:
The dissociation of $\mathrm{N}_{2} \mathrm{O}_{4}$ is- $\mathrm{N}_{2} \mathrm{O}_{4} \rightleftharpoons \quad 2 \mathrm{NO}_{2}$ Initial $\quad 1 \quad 0$ At equilibrium $1-\mathrm{x} \quad 2 \mathrm{x}$ Total numbers of moles $=1-x+2 x=1+x$ $\therefore \%$ of $\mathrm{NO}_{2}$ by Volume $=\frac{2 \mathrm{x}}{1+\mathrm{x}} \times 100=50$ $\mathrm{x}=\frac{1}{3}$ Hence, $\%$ dissociation $=33.33 \%$
BCECE-2017
Chemical Equilibrium
229101
For which among the following equimolar aqueous solutions Van't Hoff factor has the lowest value?
1 Aluminium chloride
2 Potassium sulphate
3 Ammonium chloride
4 Urea
Explanation:
For the aqueous solution of urea, Van't Hoff factor has the lowest value. Urea is non-electrolyte and does not dissociate in aqueous solution. Hence, Vant Hoff factor is 1. For Aluminium chloride $\mathrm{AlCl}_{3-}$ $\begin{gathered} \mathrm{AlCl}_{3} \rightarrow \mathrm{Al}^{3+}+3 \mathrm{Cl}^{-} \\ i=4 \end{gathered}$ For potassium sulphate $\mathrm{K}_{2} \mathrm{SO}_{4}$ $\begin{gathered} \mathrm{K}_{2} \mathrm{SO}_{4} \rightarrow 2 \mathrm{~K}^{+}+\mathrm{SO}_{4}^{2-} \\ i=3 \end{gathered}$
MHT CET-2017
Chemical Equilibrium
229103
When $\mathrm{PCl}_{5}$ is heated it gasifies and dissociates into $\mathrm{PCl}_{3}$ and $\mathrm{Cl}_{2}$. The density of the gas mixture at $200^{\circ} \mathrm{C}$ is 70.2 . What is the degree of dissociation of $\mathrm{PCl}_{5}$ at $200^{\circ} \mathrm{C}$ ?
1 0.485
2 0.242
3 0.845
4 0.542
Explanation:
The dissociation of $\mathrm{PCl}_{5}$ is$\mathrm{PCl}_{5}(\mathrm{~g}) \rightarrow \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})$ The initial vapour density will be same, it would be $\mathrm{M}_{\mathrm{PCl}_{5}} / 2$. $\therefore$ Initial vapour density $=(31+5 \times 35.5) / 2=104.25$ Vapour density at equilibrium at $200^{\circ} \mathrm{C}=70.2$ $\therefore$ Total moles at equilibrium $\begin{aligned} & (1+\alpha)=\frac{\text { Vapour density initial }}{\text { Vapour density at equilibrium }} \\ & (1+\alpha)=\frac{104.25}{70.2}=1.485 \\ & 1+\alpha=1.485 \\ & \alpha=0.485 \end{aligned}$
229098
Degree of dissociation of an acid $\mathrm{HCl}$ is $95 \%$. $0.192 \mathrm{~g}$ of the acid is present in $0.5 \mathrm{~L}$ of solution. The $\mathrm{pH}$ of the solution is :
229099
For the following equilibrium $\mathrm{N}_{2} \mathrm{O}_{4} \rightleftharpoons \quad 2 \mathrm{NO}_{2}$ in gaseous phase, $\mathrm{NO}_{2}$ is $50 \%$ of the total volume, when equilibrium is set up. Hence, percent dissociation of $\mathrm{N}_{2} \mathrm{O}_{4}$ is :
1 $50 \%$
2 $25 \%$
3 $66.66 \%$
4 $33.33 \%$
Explanation:
The dissociation of $\mathrm{N}_{2} \mathrm{O}_{4}$ is- $\mathrm{N}_{2} \mathrm{O}_{4} \rightleftharpoons \quad 2 \mathrm{NO}_{2}$ Initial $\quad 1 \quad 0$ At equilibrium $1-\mathrm{x} \quad 2 \mathrm{x}$ Total numbers of moles $=1-x+2 x=1+x$ $\therefore \%$ of $\mathrm{NO}_{2}$ by Volume $=\frac{2 \mathrm{x}}{1+\mathrm{x}} \times 100=50$ $\mathrm{x}=\frac{1}{3}$ Hence, $\%$ dissociation $=33.33 \%$
BCECE-2017
Chemical Equilibrium
229101
For which among the following equimolar aqueous solutions Van't Hoff factor has the lowest value?
1 Aluminium chloride
2 Potassium sulphate
3 Ammonium chloride
4 Urea
Explanation:
For the aqueous solution of urea, Van't Hoff factor has the lowest value. Urea is non-electrolyte and does not dissociate in aqueous solution. Hence, Vant Hoff factor is 1. For Aluminium chloride $\mathrm{AlCl}_{3-}$ $\begin{gathered} \mathrm{AlCl}_{3} \rightarrow \mathrm{Al}^{3+}+3 \mathrm{Cl}^{-} \\ i=4 \end{gathered}$ For potassium sulphate $\mathrm{K}_{2} \mathrm{SO}_{4}$ $\begin{gathered} \mathrm{K}_{2} \mathrm{SO}_{4} \rightarrow 2 \mathrm{~K}^{+}+\mathrm{SO}_{4}^{2-} \\ i=3 \end{gathered}$
MHT CET-2017
Chemical Equilibrium
229103
When $\mathrm{PCl}_{5}$ is heated it gasifies and dissociates into $\mathrm{PCl}_{3}$ and $\mathrm{Cl}_{2}$. The density of the gas mixture at $200^{\circ} \mathrm{C}$ is 70.2 . What is the degree of dissociation of $\mathrm{PCl}_{5}$ at $200^{\circ} \mathrm{C}$ ?
1 0.485
2 0.242
3 0.845
4 0.542
Explanation:
The dissociation of $\mathrm{PCl}_{5}$ is$\mathrm{PCl}_{5}(\mathrm{~g}) \rightarrow \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})$ The initial vapour density will be same, it would be $\mathrm{M}_{\mathrm{PCl}_{5}} / 2$. $\therefore$ Initial vapour density $=(31+5 \times 35.5) / 2=104.25$ Vapour density at equilibrium at $200^{\circ} \mathrm{C}=70.2$ $\therefore$ Total moles at equilibrium $\begin{aligned} & (1+\alpha)=\frac{\text { Vapour density initial }}{\text { Vapour density at equilibrium }} \\ & (1+\alpha)=\frac{104.25}{70.2}=1.485 \\ & 1+\alpha=1.485 \\ & \alpha=0.485 \end{aligned}$
229098
Degree of dissociation of an acid $\mathrm{HCl}$ is $95 \%$. $0.192 \mathrm{~g}$ of the acid is present in $0.5 \mathrm{~L}$ of solution. The $\mathrm{pH}$ of the solution is :
229099
For the following equilibrium $\mathrm{N}_{2} \mathrm{O}_{4} \rightleftharpoons \quad 2 \mathrm{NO}_{2}$ in gaseous phase, $\mathrm{NO}_{2}$ is $50 \%$ of the total volume, when equilibrium is set up. Hence, percent dissociation of $\mathrm{N}_{2} \mathrm{O}_{4}$ is :
1 $50 \%$
2 $25 \%$
3 $66.66 \%$
4 $33.33 \%$
Explanation:
The dissociation of $\mathrm{N}_{2} \mathrm{O}_{4}$ is- $\mathrm{N}_{2} \mathrm{O}_{4} \rightleftharpoons \quad 2 \mathrm{NO}_{2}$ Initial $\quad 1 \quad 0$ At equilibrium $1-\mathrm{x} \quad 2 \mathrm{x}$ Total numbers of moles $=1-x+2 x=1+x$ $\therefore \%$ of $\mathrm{NO}_{2}$ by Volume $=\frac{2 \mathrm{x}}{1+\mathrm{x}} \times 100=50$ $\mathrm{x}=\frac{1}{3}$ Hence, $\%$ dissociation $=33.33 \%$
BCECE-2017
Chemical Equilibrium
229101
For which among the following equimolar aqueous solutions Van't Hoff factor has the lowest value?
1 Aluminium chloride
2 Potassium sulphate
3 Ammonium chloride
4 Urea
Explanation:
For the aqueous solution of urea, Van't Hoff factor has the lowest value. Urea is non-electrolyte and does not dissociate in aqueous solution. Hence, Vant Hoff factor is 1. For Aluminium chloride $\mathrm{AlCl}_{3-}$ $\begin{gathered} \mathrm{AlCl}_{3} \rightarrow \mathrm{Al}^{3+}+3 \mathrm{Cl}^{-} \\ i=4 \end{gathered}$ For potassium sulphate $\mathrm{K}_{2} \mathrm{SO}_{4}$ $\begin{gathered} \mathrm{K}_{2} \mathrm{SO}_{4} \rightarrow 2 \mathrm{~K}^{+}+\mathrm{SO}_{4}^{2-} \\ i=3 \end{gathered}$
MHT CET-2017
Chemical Equilibrium
229103
When $\mathrm{PCl}_{5}$ is heated it gasifies and dissociates into $\mathrm{PCl}_{3}$ and $\mathrm{Cl}_{2}$. The density of the gas mixture at $200^{\circ} \mathrm{C}$ is 70.2 . What is the degree of dissociation of $\mathrm{PCl}_{5}$ at $200^{\circ} \mathrm{C}$ ?
1 0.485
2 0.242
3 0.845
4 0.542
Explanation:
The dissociation of $\mathrm{PCl}_{5}$ is$\mathrm{PCl}_{5}(\mathrm{~g}) \rightarrow \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})$ The initial vapour density will be same, it would be $\mathrm{M}_{\mathrm{PCl}_{5}} / 2$. $\therefore$ Initial vapour density $=(31+5 \times 35.5) / 2=104.25$ Vapour density at equilibrium at $200^{\circ} \mathrm{C}=70.2$ $\therefore$ Total moles at equilibrium $\begin{aligned} & (1+\alpha)=\frac{\text { Vapour density initial }}{\text { Vapour density at equilibrium }} \\ & (1+\alpha)=\frac{104.25}{70.2}=1.485 \\ & 1+\alpha=1.485 \\ & \alpha=0.485 \end{aligned}$
