229104
The percentage hydrolysis of 0.15 solution of ammonium acetate, $\mathrm{K}_{\mathrm{a}}$ for $\mathrm{CH}_{3} \mathrm{COOH}$ is $1.8 \times 10^{-5}$ and $\mathrm{K}_{\mathrm{b}}$ for $\mathrm{NH}_{3}$ is $\mathbf{1 . 8} \times \mathbf{1 0}^{-5}$
1 0.556
2 4.72
3 9.38
4 5.56
Explanation:
Given that, $\mathrm{K}_{\mathrm{a}}$ for $\mathrm{CH}_{3} \mathrm{COOH}=1.8 \times 10^{-5}$ $\mathrm{K}_{\mathrm{b}}$ for $\mathrm{NH}_{3}=1.8 \times 10^{-5}$ The degree of hydrolysis: $\begin{aligned} & \mathrm{h}=\sqrt{\frac{\mathrm{K}_{\mathrm{w}}}{\mathrm{K}_{\mathrm{a}} \times \mathrm{K}_{\mathrm{b}}}}=\sqrt{\frac{1 \times 10^{-14}}{1.8 \times 10^{-5} \times 1.8 \times 10^{-5}}} \\ & \mathrm{~h}=0.556 \end{aligned}$
BITSAT-2016
Chemical Equilibrium
229105
$0.6 \mathrm{~mL}$ of acetic acid is dissolved in 1 litre of water. The value of Van't Hoff factor is 1.04. What will be the degree of dissociation of the acetic acid?
4 $\quad \mathrm{Na}_{2} \mathrm{SO}_{4} \rightarrow 2 \mathrm{Na}^{+}+5 \mathrm{O}_{4}^{2-}$ $i=3$ Hence, option (a) have value of vant Hoff's factor (i) same as that of $\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}$.
Explanation:
(a) The dissociation of $\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$ is- $\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right] \Rightarrow 4 \mathrm{~K}^{+}+\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}$ Here, Van't Hoff factor $(i)=4+1=5$ (b.) $\quad \mathrm{NaCl} \rightarrow \mathrm{Na}^{+}+\mathrm{Cl}^{-}$ $i=2$ (c.) $\quad \mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3} \rightarrow \mathrm{Al}^{3+}+3 \mathrm{NO}_{3}$ $i=4$ (d.) $\quad \mathrm{Na}_{2} \mathrm{SO}_{4} \rightarrow 2 \mathrm{Na}^{+}+5 \mathrm{O}_{4}^{2-}$ $i=3$ Hence, option (a) have value of vant Hoff's factor (i) same as that of $\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}$.
COMEDK-2015
Chemical Equilibrium
229108
Van't Hoff factor of centimolal solution of $\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$ is 3.333. Calculate the per cent dissociation of $\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$.
229104
The percentage hydrolysis of 0.15 solution of ammonium acetate, $\mathrm{K}_{\mathrm{a}}$ for $\mathrm{CH}_{3} \mathrm{COOH}$ is $1.8 \times 10^{-5}$ and $\mathrm{K}_{\mathrm{b}}$ for $\mathrm{NH}_{3}$ is $\mathbf{1 . 8} \times \mathbf{1 0}^{-5}$
1 0.556
2 4.72
3 9.38
4 5.56
Explanation:
Given that, $\mathrm{K}_{\mathrm{a}}$ for $\mathrm{CH}_{3} \mathrm{COOH}=1.8 \times 10^{-5}$ $\mathrm{K}_{\mathrm{b}}$ for $\mathrm{NH}_{3}=1.8 \times 10^{-5}$ The degree of hydrolysis: $\begin{aligned} & \mathrm{h}=\sqrt{\frac{\mathrm{K}_{\mathrm{w}}}{\mathrm{K}_{\mathrm{a}} \times \mathrm{K}_{\mathrm{b}}}}=\sqrt{\frac{1 \times 10^{-14}}{1.8 \times 10^{-5} \times 1.8 \times 10^{-5}}} \\ & \mathrm{~h}=0.556 \end{aligned}$
BITSAT-2016
Chemical Equilibrium
229105
$0.6 \mathrm{~mL}$ of acetic acid is dissolved in 1 litre of water. The value of Van't Hoff factor is 1.04. What will be the degree of dissociation of the acetic acid?
4 $\quad \mathrm{Na}_{2} \mathrm{SO}_{4} \rightarrow 2 \mathrm{Na}^{+}+5 \mathrm{O}_{4}^{2-}$ $i=3$ Hence, option (a) have value of vant Hoff's factor (i) same as that of $\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}$.
Explanation:
(a) The dissociation of $\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$ is- $\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right] \Rightarrow 4 \mathrm{~K}^{+}+\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}$ Here, Van't Hoff factor $(i)=4+1=5$ (b.) $\quad \mathrm{NaCl} \rightarrow \mathrm{Na}^{+}+\mathrm{Cl}^{-}$ $i=2$ (c.) $\quad \mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3} \rightarrow \mathrm{Al}^{3+}+3 \mathrm{NO}_{3}$ $i=4$ (d.) $\quad \mathrm{Na}_{2} \mathrm{SO}_{4} \rightarrow 2 \mathrm{Na}^{+}+5 \mathrm{O}_{4}^{2-}$ $i=3$ Hence, option (a) have value of vant Hoff's factor (i) same as that of $\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}$.
COMEDK-2015
Chemical Equilibrium
229108
Van't Hoff factor of centimolal solution of $\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$ is 3.333. Calculate the per cent dissociation of $\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$.
229104
The percentage hydrolysis of 0.15 solution of ammonium acetate, $\mathrm{K}_{\mathrm{a}}$ for $\mathrm{CH}_{3} \mathrm{COOH}$ is $1.8 \times 10^{-5}$ and $\mathrm{K}_{\mathrm{b}}$ for $\mathrm{NH}_{3}$ is $\mathbf{1 . 8} \times \mathbf{1 0}^{-5}$
1 0.556
2 4.72
3 9.38
4 5.56
Explanation:
Given that, $\mathrm{K}_{\mathrm{a}}$ for $\mathrm{CH}_{3} \mathrm{COOH}=1.8 \times 10^{-5}$ $\mathrm{K}_{\mathrm{b}}$ for $\mathrm{NH}_{3}=1.8 \times 10^{-5}$ The degree of hydrolysis: $\begin{aligned} & \mathrm{h}=\sqrt{\frac{\mathrm{K}_{\mathrm{w}}}{\mathrm{K}_{\mathrm{a}} \times \mathrm{K}_{\mathrm{b}}}}=\sqrt{\frac{1 \times 10^{-14}}{1.8 \times 10^{-5} \times 1.8 \times 10^{-5}}} \\ & \mathrm{~h}=0.556 \end{aligned}$
BITSAT-2016
Chemical Equilibrium
229105
$0.6 \mathrm{~mL}$ of acetic acid is dissolved in 1 litre of water. The value of Van't Hoff factor is 1.04. What will be the degree of dissociation of the acetic acid?
4 $\quad \mathrm{Na}_{2} \mathrm{SO}_{4} \rightarrow 2 \mathrm{Na}^{+}+5 \mathrm{O}_{4}^{2-}$ $i=3$ Hence, option (a) have value of vant Hoff's factor (i) same as that of $\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}$.
Explanation:
(a) The dissociation of $\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$ is- $\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right] \Rightarrow 4 \mathrm{~K}^{+}+\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}$ Here, Van't Hoff factor $(i)=4+1=5$ (b.) $\quad \mathrm{NaCl} \rightarrow \mathrm{Na}^{+}+\mathrm{Cl}^{-}$ $i=2$ (c.) $\quad \mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3} \rightarrow \mathrm{Al}^{3+}+3 \mathrm{NO}_{3}$ $i=4$ (d.) $\quad \mathrm{Na}_{2} \mathrm{SO}_{4} \rightarrow 2 \mathrm{Na}^{+}+5 \mathrm{O}_{4}^{2-}$ $i=3$ Hence, option (a) have value of vant Hoff's factor (i) same as that of $\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}$.
COMEDK-2015
Chemical Equilibrium
229108
Van't Hoff factor of centimolal solution of $\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$ is 3.333. Calculate the per cent dissociation of $\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$.
229104
The percentage hydrolysis of 0.15 solution of ammonium acetate, $\mathrm{K}_{\mathrm{a}}$ for $\mathrm{CH}_{3} \mathrm{COOH}$ is $1.8 \times 10^{-5}$ and $\mathrm{K}_{\mathrm{b}}$ for $\mathrm{NH}_{3}$ is $\mathbf{1 . 8} \times \mathbf{1 0}^{-5}$
1 0.556
2 4.72
3 9.38
4 5.56
Explanation:
Given that, $\mathrm{K}_{\mathrm{a}}$ for $\mathrm{CH}_{3} \mathrm{COOH}=1.8 \times 10^{-5}$ $\mathrm{K}_{\mathrm{b}}$ for $\mathrm{NH}_{3}=1.8 \times 10^{-5}$ The degree of hydrolysis: $\begin{aligned} & \mathrm{h}=\sqrt{\frac{\mathrm{K}_{\mathrm{w}}}{\mathrm{K}_{\mathrm{a}} \times \mathrm{K}_{\mathrm{b}}}}=\sqrt{\frac{1 \times 10^{-14}}{1.8 \times 10^{-5} \times 1.8 \times 10^{-5}}} \\ & \mathrm{~h}=0.556 \end{aligned}$
BITSAT-2016
Chemical Equilibrium
229105
$0.6 \mathrm{~mL}$ of acetic acid is dissolved in 1 litre of water. The value of Van't Hoff factor is 1.04. What will be the degree of dissociation of the acetic acid?
4 $\quad \mathrm{Na}_{2} \mathrm{SO}_{4} \rightarrow 2 \mathrm{Na}^{+}+5 \mathrm{O}_{4}^{2-}$ $i=3$ Hence, option (a) have value of vant Hoff's factor (i) same as that of $\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}$.
Explanation:
(a) The dissociation of $\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$ is- $\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right] \Rightarrow 4 \mathrm{~K}^{+}+\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}$ Here, Van't Hoff factor $(i)=4+1=5$ (b.) $\quad \mathrm{NaCl} \rightarrow \mathrm{Na}^{+}+\mathrm{Cl}^{-}$ $i=2$ (c.) $\quad \mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3} \rightarrow \mathrm{Al}^{3+}+3 \mathrm{NO}_{3}$ $i=4$ (d.) $\quad \mathrm{Na}_{2} \mathrm{SO}_{4} \rightarrow 2 \mathrm{Na}^{+}+5 \mathrm{O}_{4}^{2-}$ $i=3$ Hence, option (a) have value of vant Hoff's factor (i) same as that of $\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}$.
COMEDK-2015
Chemical Equilibrium
229108
Van't Hoff factor of centimolal solution of $\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$ is 3.333. Calculate the per cent dissociation of $\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$.
