229080
The degree of dissociation $(\alpha)$ of a weak electrolyte, $A_{x} B_{y}$ is related to van't Hoff factor (i) by the expression
1 $\alpha=\frac{i-1}{(x+y-1)}$
2 $\alpha=\frac{i-1}{x+y-1}$
3 $\alpha=\frac{x+y-1}{i-1}$
4 $\alpha=\frac{x+y+1}{i-1}$
Explanation:
Van't Hoff factor $\begin{array}{*{35}{l}} \text{AxBy} & \rightleftharpoons & \text{x}{{\text{A}}^{\text{y}+}} & + & \text{y}{{\text{B}}^{\text{x}-}} \\ 1-\alpha & {} & \alpha x & {} & \alpha y \\ \end{array}$ Total amount of species $=1+(\mathrm{x}+\mathrm{y}-1) \alpha$ $\begin{aligned} \text { Van't Hoff factor } i & =\frac{1+(\mathrm{x}+\mathrm{y}-1) \alpha}{1} \\ \alpha & =\frac{i-1}{(\mathrm{x}+\mathrm{y}-1)} \end{aligned}$
[pTripura JEE-2022]
Chemical Equilibrium
229082
The degree of dissociated of $\mathrm{PCl}_{5}(\alpha)$ obeying the equilibrium; $\mathrm{PCl}_{5} \rightleftharpoons \mathbf{P C l}_{3}+\mathrm{Cl}_{2}$ is related to the pressure (p) at equilibrium by
1 $\alpha \propto p$
2 $\alpha \propto \frac{1}{\sqrt{p}}$
3 $\alpha \propto \frac{1}{\mathrm{p}^{2}}$
4 $\alpha \propto \frac{1}{p^{4}}$
Explanation:
-$\begin{array}{cccc} &\mathrm{PCl}_{5}(\mathrm{~g}) & \rightleftharpoons & \mathrm{PCl}_{3}(\mathrm{~g}) & + &\mathrm{Cl}_{2}(\mathrm{~g})\\ \text{Before reaction} & 1 & & 0 & & 0 \\ \text{At equilibrium} & (1- \alpha) & & \alpha & & \alpha \\ \end{array}$ Total number of moles at equilibrium $=(1-\alpha)+\alpha+\alpha=(1+\alpha)$ If $\mathrm{p}$ is the total pressure at equilibrium, then $\mathrm{P}_{\mathrm{PCl}_{5}}=\frac{(1-\alpha)}{(1+\alpha)} \times \mathrm{p}$ $\mathrm{P}_{\mathrm{PCl}_{3}}=\frac{\alpha}{(1+\alpha)} \times \mathrm{p}$ $\mathrm{P}_{\mathrm{Cl}_{2}}=\frac{\alpha}{(1+\alpha)} \times \mathrm{p}$ $\mathrm{K}_{\mathrm{p}}=\frac{\mathrm{P}_{\mathrm{PCl}_{3}} \times \mathrm{P}_{\mathrm{Cl}_{2}}}{\mathrm{P}_{\mathrm{PCl}_{5}}}$ $=\frac{[\alpha p /(1+\alpha)][\alpha p /(1+\alpha)]}{(1-\alpha) p /(1+\alpha)}$ $=\frac{\alpha^{2}}{(1+\alpha)(1-\alpha)} \times p=\frac{\alpha^{2}}{1-\alpha^{2}} \times p$ $\begin{array}{llll} \text { or } & & \alpha^{2}=\frac{\mathrm{K}_{\mathrm{p}}}{\mathrm{p}} \\ \text { or } & \alpha=\sqrt{\frac{\mathrm{K}_{\mathrm{p}}}{\mathrm{p}}} & \left(1-\alpha^{2}=1\right) \\ \text { or } & \alpha \propto \frac{1}{\sqrt{\mathrm{p}}} \end{array}$
UP CPMT-2013
Chemical Equilibrium
229083
Which one of the following electrolytes has the same value of Van't Hoff factor (i) as that of $\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$ ?
For $\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$ Since, $\mathrm{n}$ for strong electrolytes so $\mathrm{i}=5$ For, $\quad \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}$ Hence, $\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$ and $\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}$ has same Van"t Hoff factor (i)
[JIMPER-2018]
Chemical Equilibrium
229084
Which of the following acids has the smallest dissociation constant?
Lower the acidic character, smaller is the value of dissociation constant. $\text { Acidity } \propto \text { dissociation constant. }$ -I (inductive effect) of $\mathrm{F}$ is more than $\mathrm{Br}$, so $\mathrm{CH}_{3} \mathrm{CHFCOOH}$ is a stronger acid as compared to $\mathrm{CH}_{3} \mathrm{CHBrCOOH}$ As the distance between electron withdrawing group and $-\mathrm{COOH}$ group increases, acidity decreases. Thus, the order of acidity and thus, of dissociation constant is $\mathrm{BrCH}_{2} \mathrm{CH}_{2} \mathrm{COOH}<\mathrm{FCH}_{2} \mathrm{CH}_{2} \mathrm{COOH}$ (smallest $\mathrm{K}_{\mathrm{a}}$ ) $<\mathrm{CH}_{3} \mathrm{CHBrCOOH}<\mathrm{CH}_{3} \mathrm{CHFCOOH}$
NEET Test Series from KOTA - 10 Papers In MS WORD
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Chemical Equilibrium
229080
The degree of dissociation $(\alpha)$ of a weak electrolyte, $A_{x} B_{y}$ is related to van't Hoff factor (i) by the expression
1 $\alpha=\frac{i-1}{(x+y-1)}$
2 $\alpha=\frac{i-1}{x+y-1}$
3 $\alpha=\frac{x+y-1}{i-1}$
4 $\alpha=\frac{x+y+1}{i-1}$
Explanation:
Van't Hoff factor $\begin{array}{*{35}{l}} \text{AxBy} & \rightleftharpoons & \text{x}{{\text{A}}^{\text{y}+}} & + & \text{y}{{\text{B}}^{\text{x}-}} \\ 1-\alpha & {} & \alpha x & {} & \alpha y \\ \end{array}$ Total amount of species $=1+(\mathrm{x}+\mathrm{y}-1) \alpha$ $\begin{aligned} \text { Van't Hoff factor } i & =\frac{1+(\mathrm{x}+\mathrm{y}-1) \alpha}{1} \\ \alpha & =\frac{i-1}{(\mathrm{x}+\mathrm{y}-1)} \end{aligned}$
[pTripura JEE-2022]
Chemical Equilibrium
229082
The degree of dissociated of $\mathrm{PCl}_{5}(\alpha)$ obeying the equilibrium; $\mathrm{PCl}_{5} \rightleftharpoons \mathbf{P C l}_{3}+\mathrm{Cl}_{2}$ is related to the pressure (p) at equilibrium by
1 $\alpha \propto p$
2 $\alpha \propto \frac{1}{\sqrt{p}}$
3 $\alpha \propto \frac{1}{\mathrm{p}^{2}}$
4 $\alpha \propto \frac{1}{p^{4}}$
Explanation:
-$\begin{array}{cccc} &\mathrm{PCl}_{5}(\mathrm{~g}) & \rightleftharpoons & \mathrm{PCl}_{3}(\mathrm{~g}) & + &\mathrm{Cl}_{2}(\mathrm{~g})\\ \text{Before reaction} & 1 & & 0 & & 0 \\ \text{At equilibrium} & (1- \alpha) & & \alpha & & \alpha \\ \end{array}$ Total number of moles at equilibrium $=(1-\alpha)+\alpha+\alpha=(1+\alpha)$ If $\mathrm{p}$ is the total pressure at equilibrium, then $\mathrm{P}_{\mathrm{PCl}_{5}}=\frac{(1-\alpha)}{(1+\alpha)} \times \mathrm{p}$ $\mathrm{P}_{\mathrm{PCl}_{3}}=\frac{\alpha}{(1+\alpha)} \times \mathrm{p}$ $\mathrm{P}_{\mathrm{Cl}_{2}}=\frac{\alpha}{(1+\alpha)} \times \mathrm{p}$ $\mathrm{K}_{\mathrm{p}}=\frac{\mathrm{P}_{\mathrm{PCl}_{3}} \times \mathrm{P}_{\mathrm{Cl}_{2}}}{\mathrm{P}_{\mathrm{PCl}_{5}}}$ $=\frac{[\alpha p /(1+\alpha)][\alpha p /(1+\alpha)]}{(1-\alpha) p /(1+\alpha)}$ $=\frac{\alpha^{2}}{(1+\alpha)(1-\alpha)} \times p=\frac{\alpha^{2}}{1-\alpha^{2}} \times p$ $\begin{array}{llll} \text { or } & & \alpha^{2}=\frac{\mathrm{K}_{\mathrm{p}}}{\mathrm{p}} \\ \text { or } & \alpha=\sqrt{\frac{\mathrm{K}_{\mathrm{p}}}{\mathrm{p}}} & \left(1-\alpha^{2}=1\right) \\ \text { or } & \alpha \propto \frac{1}{\sqrt{\mathrm{p}}} \end{array}$
UP CPMT-2013
Chemical Equilibrium
229083
Which one of the following electrolytes has the same value of Van't Hoff factor (i) as that of $\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$ ?
For $\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$ Since, $\mathrm{n}$ for strong electrolytes so $\mathrm{i}=5$ For, $\quad \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}$ Hence, $\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$ and $\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}$ has same Van"t Hoff factor (i)
[JIMPER-2018]
Chemical Equilibrium
229084
Which of the following acids has the smallest dissociation constant?
Lower the acidic character, smaller is the value of dissociation constant. $\text { Acidity } \propto \text { dissociation constant. }$ -I (inductive effect) of $\mathrm{F}$ is more than $\mathrm{Br}$, so $\mathrm{CH}_{3} \mathrm{CHFCOOH}$ is a stronger acid as compared to $\mathrm{CH}_{3} \mathrm{CHBrCOOH}$ As the distance between electron withdrawing group and $-\mathrm{COOH}$ group increases, acidity decreases. Thus, the order of acidity and thus, of dissociation constant is $\mathrm{BrCH}_{2} \mathrm{CH}_{2} \mathrm{COOH}<\mathrm{FCH}_{2} \mathrm{CH}_{2} \mathrm{COOH}$ (smallest $\mathrm{K}_{\mathrm{a}}$ ) $<\mathrm{CH}_{3} \mathrm{CHBrCOOH}<\mathrm{CH}_{3} \mathrm{CHFCOOH}$
229080
The degree of dissociation $(\alpha)$ of a weak electrolyte, $A_{x} B_{y}$ is related to van't Hoff factor (i) by the expression
1 $\alpha=\frac{i-1}{(x+y-1)}$
2 $\alpha=\frac{i-1}{x+y-1}$
3 $\alpha=\frac{x+y-1}{i-1}$
4 $\alpha=\frac{x+y+1}{i-1}$
Explanation:
Van't Hoff factor $\begin{array}{*{35}{l}} \text{AxBy} & \rightleftharpoons & \text{x}{{\text{A}}^{\text{y}+}} & + & \text{y}{{\text{B}}^{\text{x}-}} \\ 1-\alpha & {} & \alpha x & {} & \alpha y \\ \end{array}$ Total amount of species $=1+(\mathrm{x}+\mathrm{y}-1) \alpha$ $\begin{aligned} \text { Van't Hoff factor } i & =\frac{1+(\mathrm{x}+\mathrm{y}-1) \alpha}{1} \\ \alpha & =\frac{i-1}{(\mathrm{x}+\mathrm{y}-1)} \end{aligned}$
[pTripura JEE-2022]
Chemical Equilibrium
229082
The degree of dissociated of $\mathrm{PCl}_{5}(\alpha)$ obeying the equilibrium; $\mathrm{PCl}_{5} \rightleftharpoons \mathbf{P C l}_{3}+\mathrm{Cl}_{2}$ is related to the pressure (p) at equilibrium by
1 $\alpha \propto p$
2 $\alpha \propto \frac{1}{\sqrt{p}}$
3 $\alpha \propto \frac{1}{\mathrm{p}^{2}}$
4 $\alpha \propto \frac{1}{p^{4}}$
Explanation:
-$\begin{array}{cccc} &\mathrm{PCl}_{5}(\mathrm{~g}) & \rightleftharpoons & \mathrm{PCl}_{3}(\mathrm{~g}) & + &\mathrm{Cl}_{2}(\mathrm{~g})\\ \text{Before reaction} & 1 & & 0 & & 0 \\ \text{At equilibrium} & (1- \alpha) & & \alpha & & \alpha \\ \end{array}$ Total number of moles at equilibrium $=(1-\alpha)+\alpha+\alpha=(1+\alpha)$ If $\mathrm{p}$ is the total pressure at equilibrium, then $\mathrm{P}_{\mathrm{PCl}_{5}}=\frac{(1-\alpha)}{(1+\alpha)} \times \mathrm{p}$ $\mathrm{P}_{\mathrm{PCl}_{3}}=\frac{\alpha}{(1+\alpha)} \times \mathrm{p}$ $\mathrm{P}_{\mathrm{Cl}_{2}}=\frac{\alpha}{(1+\alpha)} \times \mathrm{p}$ $\mathrm{K}_{\mathrm{p}}=\frac{\mathrm{P}_{\mathrm{PCl}_{3}} \times \mathrm{P}_{\mathrm{Cl}_{2}}}{\mathrm{P}_{\mathrm{PCl}_{5}}}$ $=\frac{[\alpha p /(1+\alpha)][\alpha p /(1+\alpha)]}{(1-\alpha) p /(1+\alpha)}$ $=\frac{\alpha^{2}}{(1+\alpha)(1-\alpha)} \times p=\frac{\alpha^{2}}{1-\alpha^{2}} \times p$ $\begin{array}{llll} \text { or } & & \alpha^{2}=\frac{\mathrm{K}_{\mathrm{p}}}{\mathrm{p}} \\ \text { or } & \alpha=\sqrt{\frac{\mathrm{K}_{\mathrm{p}}}{\mathrm{p}}} & \left(1-\alpha^{2}=1\right) \\ \text { or } & \alpha \propto \frac{1}{\sqrt{\mathrm{p}}} \end{array}$
UP CPMT-2013
Chemical Equilibrium
229083
Which one of the following electrolytes has the same value of Van't Hoff factor (i) as that of $\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$ ?
For $\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$ Since, $\mathrm{n}$ for strong electrolytes so $\mathrm{i}=5$ For, $\quad \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}$ Hence, $\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$ and $\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}$ has same Van"t Hoff factor (i)
[JIMPER-2018]
Chemical Equilibrium
229084
Which of the following acids has the smallest dissociation constant?
Lower the acidic character, smaller is the value of dissociation constant. $\text { Acidity } \propto \text { dissociation constant. }$ -I (inductive effect) of $\mathrm{F}$ is more than $\mathrm{Br}$, so $\mathrm{CH}_{3} \mathrm{CHFCOOH}$ is a stronger acid as compared to $\mathrm{CH}_{3} \mathrm{CHBrCOOH}$ As the distance between electron withdrawing group and $-\mathrm{COOH}$ group increases, acidity decreases. Thus, the order of acidity and thus, of dissociation constant is $\mathrm{BrCH}_{2} \mathrm{CH}_{2} \mathrm{COOH}<\mathrm{FCH}_{2} \mathrm{CH}_{2} \mathrm{COOH}$ (smallest $\mathrm{K}_{\mathrm{a}}$ ) $<\mathrm{CH}_{3} \mathrm{CHBrCOOH}<\mathrm{CH}_{3} \mathrm{CHFCOOH}$
229080
The degree of dissociation $(\alpha)$ of a weak electrolyte, $A_{x} B_{y}$ is related to van't Hoff factor (i) by the expression
1 $\alpha=\frac{i-1}{(x+y-1)}$
2 $\alpha=\frac{i-1}{x+y-1}$
3 $\alpha=\frac{x+y-1}{i-1}$
4 $\alpha=\frac{x+y+1}{i-1}$
Explanation:
Van't Hoff factor $\begin{array}{*{35}{l}} \text{AxBy} & \rightleftharpoons & \text{x}{{\text{A}}^{\text{y}+}} & + & \text{y}{{\text{B}}^{\text{x}-}} \\ 1-\alpha & {} & \alpha x & {} & \alpha y \\ \end{array}$ Total amount of species $=1+(\mathrm{x}+\mathrm{y}-1) \alpha$ $\begin{aligned} \text { Van't Hoff factor } i & =\frac{1+(\mathrm{x}+\mathrm{y}-1) \alpha}{1} \\ \alpha & =\frac{i-1}{(\mathrm{x}+\mathrm{y}-1)} \end{aligned}$
[pTripura JEE-2022]
Chemical Equilibrium
229082
The degree of dissociated of $\mathrm{PCl}_{5}(\alpha)$ obeying the equilibrium; $\mathrm{PCl}_{5} \rightleftharpoons \mathbf{P C l}_{3}+\mathrm{Cl}_{2}$ is related to the pressure (p) at equilibrium by
1 $\alpha \propto p$
2 $\alpha \propto \frac{1}{\sqrt{p}}$
3 $\alpha \propto \frac{1}{\mathrm{p}^{2}}$
4 $\alpha \propto \frac{1}{p^{4}}$
Explanation:
-$\begin{array}{cccc} &\mathrm{PCl}_{5}(\mathrm{~g}) & \rightleftharpoons & \mathrm{PCl}_{3}(\mathrm{~g}) & + &\mathrm{Cl}_{2}(\mathrm{~g})\\ \text{Before reaction} & 1 & & 0 & & 0 \\ \text{At equilibrium} & (1- \alpha) & & \alpha & & \alpha \\ \end{array}$ Total number of moles at equilibrium $=(1-\alpha)+\alpha+\alpha=(1+\alpha)$ If $\mathrm{p}$ is the total pressure at equilibrium, then $\mathrm{P}_{\mathrm{PCl}_{5}}=\frac{(1-\alpha)}{(1+\alpha)} \times \mathrm{p}$ $\mathrm{P}_{\mathrm{PCl}_{3}}=\frac{\alpha}{(1+\alpha)} \times \mathrm{p}$ $\mathrm{P}_{\mathrm{Cl}_{2}}=\frac{\alpha}{(1+\alpha)} \times \mathrm{p}$ $\mathrm{K}_{\mathrm{p}}=\frac{\mathrm{P}_{\mathrm{PCl}_{3}} \times \mathrm{P}_{\mathrm{Cl}_{2}}}{\mathrm{P}_{\mathrm{PCl}_{5}}}$ $=\frac{[\alpha p /(1+\alpha)][\alpha p /(1+\alpha)]}{(1-\alpha) p /(1+\alpha)}$ $=\frac{\alpha^{2}}{(1+\alpha)(1-\alpha)} \times p=\frac{\alpha^{2}}{1-\alpha^{2}} \times p$ $\begin{array}{llll} \text { or } & & \alpha^{2}=\frac{\mathrm{K}_{\mathrm{p}}}{\mathrm{p}} \\ \text { or } & \alpha=\sqrt{\frac{\mathrm{K}_{\mathrm{p}}}{\mathrm{p}}} & \left(1-\alpha^{2}=1\right) \\ \text { or } & \alpha \propto \frac{1}{\sqrt{\mathrm{p}}} \end{array}$
UP CPMT-2013
Chemical Equilibrium
229083
Which one of the following electrolytes has the same value of Van't Hoff factor (i) as that of $\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$ ?
For $\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$ Since, $\mathrm{n}$ for strong electrolytes so $\mathrm{i}=5$ For, $\quad \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}$ Hence, $\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$ and $\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}$ has same Van"t Hoff factor (i)
[JIMPER-2018]
Chemical Equilibrium
229084
Which of the following acids has the smallest dissociation constant?
Lower the acidic character, smaller is the value of dissociation constant. $\text { Acidity } \propto \text { dissociation constant. }$ -I (inductive effect) of $\mathrm{F}$ is more than $\mathrm{Br}$, so $\mathrm{CH}_{3} \mathrm{CHFCOOH}$ is a stronger acid as compared to $\mathrm{CH}_{3} \mathrm{CHBrCOOH}$ As the distance between electron withdrawing group and $-\mathrm{COOH}$ group increases, acidity decreases. Thus, the order of acidity and thus, of dissociation constant is $\mathrm{BrCH}_{2} \mathrm{CH}_{2} \mathrm{COOH}<\mathrm{FCH}_{2} \mathrm{CH}_{2} \mathrm{COOH}$ (smallest $\mathrm{K}_{\mathrm{a}}$ ) $<\mathrm{CH}_{3} \mathrm{CHBrCOOH}<\mathrm{CH}_{3} \mathrm{CHFCOOH}$
