229132
A monoprotic acid in $1.00 \mathrm{M}$ solution is $\mathbf{0 . 0 1 \%}$ ionised. The dissociation constant of this acid is
1 $1 \times 10^{-8}$
2 $1 \times 10^{-4}$
3 $1 \times 10^{-6}$
4 $1 \times 10^{-5}$
Explanation:
Given, $\mathrm{C}=1.00 \mathrm{M}, \mathrm{K}_{\mathrm{a}}=$ ? We use the Ostwald's dilution law for weak electrolytes- $\begin{aligned} \alpha & =\sqrt{\mathrm{K}_{\mathrm{a}} \mathrm{V}} \\ \text { Here, } \quad \alpha & =0.001 \%=0.0001=1 \times 10^{-4} \\ \mathrm{~V} & =\frac{1}{\mathrm{C}}=\frac{1}{1-0}=1 \mathrm{~L} \\ \therefore \quad \mathrm{K}_{\mathrm{a}} & =\frac{\alpha^2}{\mathrm{~V}}=\frac{\left(1 \times 10^{-4}\right)^2}{1} \\ & =1 \times 10^{-8}\end{aligned}$
BCECE-2006
Chemical Equilibrium
229136
If $\alpha$ is the degree of dissociation of $\mathrm{Na}_{2} \mathrm{SO}_{4}$, the van't Hoff factor $(i)$ used for calculating the molecular mass is
1 $1-2 \alpha$
2 $1+2 \alpha$
3 $1-\alpha$
4 $1+\alpha$
Explanation:
The dissociation of $\mathrm{Na}_{2} \mathrm{SO}_{4}$ is given as- $\mathrm{Na}_{2} \mathrm{SO}_{4} \longrightarrow \underbrace{2 \mathrm{Na}^{+}+\mathrm{SO}_{4}^{-2}}_{\mathrm{n}=2}$ $\therefore \quad i=1+(\mathrm{n}-1) \alpha$ $i=1+2 \alpha$
AIEEE-2005
Chemical Equilibrium
229138
When $\mathrm{NH}_{4} \mathrm{Cl}$ is added to $\mathrm{NH}_{4} \mathrm{OH}$ solution, the dissociation of ammonium hydroxide is reduced. it is due to
1 common ion effect
2 hydrolysis
3 oxidation
4 reduction
Explanation:
When $\mathrm{NH}_{4} \mathrm{Cl}$ is added to $\mathrm{NH}_{4} \mathrm{OH}$ solution, concentration of $\mathrm{NH}_{4}{ }^{+}$ions increases so the equilibrium shift towards left. So, the dissociation of $\mathrm{NH}_{4} \mathrm{OH}$ is reduced it is due to common ion effect.
CG PET-2004
Chemical Equilibrium
229143
Which of the following salt has value of the van't Hoff factor equal to that of $\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$ ?
$\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$ on ionisation gives 4 species $\left(3 \mathrm{~K}^{+}\right.$and $\left.1\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{-3}\right)$. And, $\quad \mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}$ also gives 4 species $\left(1 \mathrm{Al}^{3+}\right.$ and $3 \mathrm{NO}_{3}^{-}$). $\therefore \quad$ van't Hoff factor for $\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]=4$ And, van't Hoff factor for $\mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}=4$
229132
A monoprotic acid in $1.00 \mathrm{M}$ solution is $\mathbf{0 . 0 1 \%}$ ionised. The dissociation constant of this acid is
1 $1 \times 10^{-8}$
2 $1 \times 10^{-4}$
3 $1 \times 10^{-6}$
4 $1 \times 10^{-5}$
Explanation:
Given, $\mathrm{C}=1.00 \mathrm{M}, \mathrm{K}_{\mathrm{a}}=$ ? We use the Ostwald's dilution law for weak electrolytes- $\begin{aligned} \alpha & =\sqrt{\mathrm{K}_{\mathrm{a}} \mathrm{V}} \\ \text { Here, } \quad \alpha & =0.001 \%=0.0001=1 \times 10^{-4} \\ \mathrm{~V} & =\frac{1}{\mathrm{C}}=\frac{1}{1-0}=1 \mathrm{~L} \\ \therefore \quad \mathrm{K}_{\mathrm{a}} & =\frac{\alpha^2}{\mathrm{~V}}=\frac{\left(1 \times 10^{-4}\right)^2}{1} \\ & =1 \times 10^{-8}\end{aligned}$
BCECE-2006
Chemical Equilibrium
229136
If $\alpha$ is the degree of dissociation of $\mathrm{Na}_{2} \mathrm{SO}_{4}$, the van't Hoff factor $(i)$ used for calculating the molecular mass is
1 $1-2 \alpha$
2 $1+2 \alpha$
3 $1-\alpha$
4 $1+\alpha$
Explanation:
The dissociation of $\mathrm{Na}_{2} \mathrm{SO}_{4}$ is given as- $\mathrm{Na}_{2} \mathrm{SO}_{4} \longrightarrow \underbrace{2 \mathrm{Na}^{+}+\mathrm{SO}_{4}^{-2}}_{\mathrm{n}=2}$ $\therefore \quad i=1+(\mathrm{n}-1) \alpha$ $i=1+2 \alpha$
AIEEE-2005
Chemical Equilibrium
229138
When $\mathrm{NH}_{4} \mathrm{Cl}$ is added to $\mathrm{NH}_{4} \mathrm{OH}$ solution, the dissociation of ammonium hydroxide is reduced. it is due to
1 common ion effect
2 hydrolysis
3 oxidation
4 reduction
Explanation:
When $\mathrm{NH}_{4} \mathrm{Cl}$ is added to $\mathrm{NH}_{4} \mathrm{OH}$ solution, concentration of $\mathrm{NH}_{4}{ }^{+}$ions increases so the equilibrium shift towards left. So, the dissociation of $\mathrm{NH}_{4} \mathrm{OH}$ is reduced it is due to common ion effect.
CG PET-2004
Chemical Equilibrium
229143
Which of the following salt has value of the van't Hoff factor equal to that of $\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$ ?
$\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$ on ionisation gives 4 species $\left(3 \mathrm{~K}^{+}\right.$and $\left.1\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{-3}\right)$. And, $\quad \mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}$ also gives 4 species $\left(1 \mathrm{Al}^{3+}\right.$ and $3 \mathrm{NO}_{3}^{-}$). $\therefore \quad$ van't Hoff factor for $\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]=4$ And, van't Hoff factor for $\mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}=4$
229132
A monoprotic acid in $1.00 \mathrm{M}$ solution is $\mathbf{0 . 0 1 \%}$ ionised. The dissociation constant of this acid is
1 $1 \times 10^{-8}$
2 $1 \times 10^{-4}$
3 $1 \times 10^{-6}$
4 $1 \times 10^{-5}$
Explanation:
Given, $\mathrm{C}=1.00 \mathrm{M}, \mathrm{K}_{\mathrm{a}}=$ ? We use the Ostwald's dilution law for weak electrolytes- $\begin{aligned} \alpha & =\sqrt{\mathrm{K}_{\mathrm{a}} \mathrm{V}} \\ \text { Here, } \quad \alpha & =0.001 \%=0.0001=1 \times 10^{-4} \\ \mathrm{~V} & =\frac{1}{\mathrm{C}}=\frac{1}{1-0}=1 \mathrm{~L} \\ \therefore \quad \mathrm{K}_{\mathrm{a}} & =\frac{\alpha^2}{\mathrm{~V}}=\frac{\left(1 \times 10^{-4}\right)^2}{1} \\ & =1 \times 10^{-8}\end{aligned}$
BCECE-2006
Chemical Equilibrium
229136
If $\alpha$ is the degree of dissociation of $\mathrm{Na}_{2} \mathrm{SO}_{4}$, the van't Hoff factor $(i)$ used for calculating the molecular mass is
1 $1-2 \alpha$
2 $1+2 \alpha$
3 $1-\alpha$
4 $1+\alpha$
Explanation:
The dissociation of $\mathrm{Na}_{2} \mathrm{SO}_{4}$ is given as- $\mathrm{Na}_{2} \mathrm{SO}_{4} \longrightarrow \underbrace{2 \mathrm{Na}^{+}+\mathrm{SO}_{4}^{-2}}_{\mathrm{n}=2}$ $\therefore \quad i=1+(\mathrm{n}-1) \alpha$ $i=1+2 \alpha$
AIEEE-2005
Chemical Equilibrium
229138
When $\mathrm{NH}_{4} \mathrm{Cl}$ is added to $\mathrm{NH}_{4} \mathrm{OH}$ solution, the dissociation of ammonium hydroxide is reduced. it is due to
1 common ion effect
2 hydrolysis
3 oxidation
4 reduction
Explanation:
When $\mathrm{NH}_{4} \mathrm{Cl}$ is added to $\mathrm{NH}_{4} \mathrm{OH}$ solution, concentration of $\mathrm{NH}_{4}{ }^{+}$ions increases so the equilibrium shift towards left. So, the dissociation of $\mathrm{NH}_{4} \mathrm{OH}$ is reduced it is due to common ion effect.
CG PET-2004
Chemical Equilibrium
229143
Which of the following salt has value of the van't Hoff factor equal to that of $\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$ ?
$\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$ on ionisation gives 4 species $\left(3 \mathrm{~K}^{+}\right.$and $\left.1\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{-3}\right)$. And, $\quad \mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}$ also gives 4 species $\left(1 \mathrm{Al}^{3+}\right.$ and $3 \mathrm{NO}_{3}^{-}$). $\therefore \quad$ van't Hoff factor for $\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]=4$ And, van't Hoff factor for $\mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}=4$
229132
A monoprotic acid in $1.00 \mathrm{M}$ solution is $\mathbf{0 . 0 1 \%}$ ionised. The dissociation constant of this acid is
1 $1 \times 10^{-8}$
2 $1 \times 10^{-4}$
3 $1 \times 10^{-6}$
4 $1 \times 10^{-5}$
Explanation:
Given, $\mathrm{C}=1.00 \mathrm{M}, \mathrm{K}_{\mathrm{a}}=$ ? We use the Ostwald's dilution law for weak electrolytes- $\begin{aligned} \alpha & =\sqrt{\mathrm{K}_{\mathrm{a}} \mathrm{V}} \\ \text { Here, } \quad \alpha & =0.001 \%=0.0001=1 \times 10^{-4} \\ \mathrm{~V} & =\frac{1}{\mathrm{C}}=\frac{1}{1-0}=1 \mathrm{~L} \\ \therefore \quad \mathrm{K}_{\mathrm{a}} & =\frac{\alpha^2}{\mathrm{~V}}=\frac{\left(1 \times 10^{-4}\right)^2}{1} \\ & =1 \times 10^{-8}\end{aligned}$
BCECE-2006
Chemical Equilibrium
229136
If $\alpha$ is the degree of dissociation of $\mathrm{Na}_{2} \mathrm{SO}_{4}$, the van't Hoff factor $(i)$ used for calculating the molecular mass is
1 $1-2 \alpha$
2 $1+2 \alpha$
3 $1-\alpha$
4 $1+\alpha$
Explanation:
The dissociation of $\mathrm{Na}_{2} \mathrm{SO}_{4}$ is given as- $\mathrm{Na}_{2} \mathrm{SO}_{4} \longrightarrow \underbrace{2 \mathrm{Na}^{+}+\mathrm{SO}_{4}^{-2}}_{\mathrm{n}=2}$ $\therefore \quad i=1+(\mathrm{n}-1) \alpha$ $i=1+2 \alpha$
AIEEE-2005
Chemical Equilibrium
229138
When $\mathrm{NH}_{4} \mathrm{Cl}$ is added to $\mathrm{NH}_{4} \mathrm{OH}$ solution, the dissociation of ammonium hydroxide is reduced. it is due to
1 common ion effect
2 hydrolysis
3 oxidation
4 reduction
Explanation:
When $\mathrm{NH}_{4} \mathrm{Cl}$ is added to $\mathrm{NH}_{4} \mathrm{OH}$ solution, concentration of $\mathrm{NH}_{4}{ }^{+}$ions increases so the equilibrium shift towards left. So, the dissociation of $\mathrm{NH}_{4} \mathrm{OH}$ is reduced it is due to common ion effect.
CG PET-2004
Chemical Equilibrium
229143
Which of the following salt has value of the van't Hoff factor equal to that of $\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$ ?
$\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$ on ionisation gives 4 species $\left(3 \mathrm{~K}^{+}\right.$and $\left.1\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{-3}\right)$. And, $\quad \mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}$ also gives 4 species $\left(1 \mathrm{Al}^{3+}\right.$ and $3 \mathrm{NO}_{3}^{-}$). $\therefore \quad$ van't Hoff factor for $\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]=4$ And, van't Hoff factor for $\mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}=4$
