229078
Van't Hoff factor of $\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}$ is
1 One
2 Two
3 Three
4 Four
Explanation:
$\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2} \rightleftharpoons \mathrm{Ca}^{2+}+2 \mathrm{NO}_{3}^{-}$ It furnishes 3 ions per formula unit. So, its Van't Hoff factor is 3.
[J and K CET-(2013)]
Chemical Equilibrium
229095
$\mathrm{K}_{2} \mathrm{HgI}_{4}$ is $40 \%$ ionised in aqueous solution. The value of its Van't Hoff factor (i) is
1 1.6
2 1.8
3 2.2
4 2.0
Explanation:
The ionisation of $\mathrm{K}_{2} \mathrm{HgI}_{4}$ is given as$\mathrm{K}_{2} \mathrm{HgI}_{4} \longrightarrow 2 \mathrm{~K}^{+}+\mathrm{HgI}_{4}^{2-}$ $i=1+(\mathrm{n}-1) \alpha$ $\alpha=0.4, \mathrm{n}=3$ So, $\quad i=1+2 \times 0.4=1.8$
[JEE Mains 2019
Chemical Equilibrium
229121
If $\mathrm{BaCl}_{2}$ ionizes to an extent of $80 \%$ in aqueous solution, the value of van't Hoff factor is
1 2.6
2 0.4
3 0.8
4 2.4
Explanation:
Ionisation of $\mathrm{BaCl}_{2}$ is-
AP-EAMCET- (Engg.) - 2010
Chemical Equilibrium
229123
Van't Hoff factor of aq $\mathrm{K}_{2} \mathrm{SO}_{4}$ at infinite dilution has value equal to
1 1
2 2
3 3
4 between 2 and 3
Explanation:
At infinite dilution, complete dissociation will take place. So, $\mathrm{K}_{2} \mathrm{SO}_{4} \rightarrow 2 \mathrm{~K}^{+}+\mathrm{SO}_{4}{ }^{2-}$ van't Hoff factor $=2+1=3$
AMU-2009
Chemical Equilibrium
229128
In a $500 \mathrm{~mL}$ flask, the degree of dissociation of $\mathrm{PCl}_{5}$ at equilibrium is $40 \%$ and the initial amount is 5 moles. The value of equilibrium constant in mol $\mathrm{L}^{-1}$ for the decomposition of $\mathrm{PCl}_{5}$ is
229078
Van't Hoff factor of $\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}$ is
1 One
2 Two
3 Three
4 Four
Explanation:
$\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2} \rightleftharpoons \mathrm{Ca}^{2+}+2 \mathrm{NO}_{3}^{-}$ It furnishes 3 ions per formula unit. So, its Van't Hoff factor is 3.
[J and K CET-(2013)]
Chemical Equilibrium
229095
$\mathrm{K}_{2} \mathrm{HgI}_{4}$ is $40 \%$ ionised in aqueous solution. The value of its Van't Hoff factor (i) is
1 1.6
2 1.8
3 2.2
4 2.0
Explanation:
The ionisation of $\mathrm{K}_{2} \mathrm{HgI}_{4}$ is given as$\mathrm{K}_{2} \mathrm{HgI}_{4} \longrightarrow 2 \mathrm{~K}^{+}+\mathrm{HgI}_{4}^{2-}$ $i=1+(\mathrm{n}-1) \alpha$ $\alpha=0.4, \mathrm{n}=3$ So, $\quad i=1+2 \times 0.4=1.8$
[JEE Mains 2019
Chemical Equilibrium
229121
If $\mathrm{BaCl}_{2}$ ionizes to an extent of $80 \%$ in aqueous solution, the value of van't Hoff factor is
1 2.6
2 0.4
3 0.8
4 2.4
Explanation:
Ionisation of $\mathrm{BaCl}_{2}$ is-
AP-EAMCET- (Engg.) - 2010
Chemical Equilibrium
229123
Van't Hoff factor of aq $\mathrm{K}_{2} \mathrm{SO}_{4}$ at infinite dilution has value equal to
1 1
2 2
3 3
4 between 2 and 3
Explanation:
At infinite dilution, complete dissociation will take place. So, $\mathrm{K}_{2} \mathrm{SO}_{4} \rightarrow 2 \mathrm{~K}^{+}+\mathrm{SO}_{4}{ }^{2-}$ van't Hoff factor $=2+1=3$
AMU-2009
Chemical Equilibrium
229128
In a $500 \mathrm{~mL}$ flask, the degree of dissociation of $\mathrm{PCl}_{5}$ at equilibrium is $40 \%$ and the initial amount is 5 moles. The value of equilibrium constant in mol $\mathrm{L}^{-1}$ for the decomposition of $\mathrm{PCl}_{5}$ is
229078
Van't Hoff factor of $\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}$ is
1 One
2 Two
3 Three
4 Four
Explanation:
$\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2} \rightleftharpoons \mathrm{Ca}^{2+}+2 \mathrm{NO}_{3}^{-}$ It furnishes 3 ions per formula unit. So, its Van't Hoff factor is 3.
[J and K CET-(2013)]
Chemical Equilibrium
229095
$\mathrm{K}_{2} \mathrm{HgI}_{4}$ is $40 \%$ ionised in aqueous solution. The value of its Van't Hoff factor (i) is
1 1.6
2 1.8
3 2.2
4 2.0
Explanation:
The ionisation of $\mathrm{K}_{2} \mathrm{HgI}_{4}$ is given as$\mathrm{K}_{2} \mathrm{HgI}_{4} \longrightarrow 2 \mathrm{~K}^{+}+\mathrm{HgI}_{4}^{2-}$ $i=1+(\mathrm{n}-1) \alpha$ $\alpha=0.4, \mathrm{n}=3$ So, $\quad i=1+2 \times 0.4=1.8$
[JEE Mains 2019
Chemical Equilibrium
229121
If $\mathrm{BaCl}_{2}$ ionizes to an extent of $80 \%$ in aqueous solution, the value of van't Hoff factor is
1 2.6
2 0.4
3 0.8
4 2.4
Explanation:
Ionisation of $\mathrm{BaCl}_{2}$ is-
AP-EAMCET- (Engg.) - 2010
Chemical Equilibrium
229123
Van't Hoff factor of aq $\mathrm{K}_{2} \mathrm{SO}_{4}$ at infinite dilution has value equal to
1 1
2 2
3 3
4 between 2 and 3
Explanation:
At infinite dilution, complete dissociation will take place. So, $\mathrm{K}_{2} \mathrm{SO}_{4} \rightarrow 2 \mathrm{~K}^{+}+\mathrm{SO}_{4}{ }^{2-}$ van't Hoff factor $=2+1=3$
AMU-2009
Chemical Equilibrium
229128
In a $500 \mathrm{~mL}$ flask, the degree of dissociation of $\mathrm{PCl}_{5}$ at equilibrium is $40 \%$ and the initial amount is 5 moles. The value of equilibrium constant in mol $\mathrm{L}^{-1}$ for the decomposition of $\mathrm{PCl}_{5}$ is
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Chemical Equilibrium
229078
Van't Hoff factor of $\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}$ is
1 One
2 Two
3 Three
4 Four
Explanation:
$\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2} \rightleftharpoons \mathrm{Ca}^{2+}+2 \mathrm{NO}_{3}^{-}$ It furnishes 3 ions per formula unit. So, its Van't Hoff factor is 3.
[J and K CET-(2013)]
Chemical Equilibrium
229095
$\mathrm{K}_{2} \mathrm{HgI}_{4}$ is $40 \%$ ionised in aqueous solution. The value of its Van't Hoff factor (i) is
1 1.6
2 1.8
3 2.2
4 2.0
Explanation:
The ionisation of $\mathrm{K}_{2} \mathrm{HgI}_{4}$ is given as$\mathrm{K}_{2} \mathrm{HgI}_{4} \longrightarrow 2 \mathrm{~K}^{+}+\mathrm{HgI}_{4}^{2-}$ $i=1+(\mathrm{n}-1) \alpha$ $\alpha=0.4, \mathrm{n}=3$ So, $\quad i=1+2 \times 0.4=1.8$
[JEE Mains 2019
Chemical Equilibrium
229121
If $\mathrm{BaCl}_{2}$ ionizes to an extent of $80 \%$ in aqueous solution, the value of van't Hoff factor is
1 2.6
2 0.4
3 0.8
4 2.4
Explanation:
Ionisation of $\mathrm{BaCl}_{2}$ is-
AP-EAMCET- (Engg.) - 2010
Chemical Equilibrium
229123
Van't Hoff factor of aq $\mathrm{K}_{2} \mathrm{SO}_{4}$ at infinite dilution has value equal to
1 1
2 2
3 3
4 between 2 and 3
Explanation:
At infinite dilution, complete dissociation will take place. So, $\mathrm{K}_{2} \mathrm{SO}_{4} \rightarrow 2 \mathrm{~K}^{+}+\mathrm{SO}_{4}{ }^{2-}$ van't Hoff factor $=2+1=3$
AMU-2009
Chemical Equilibrium
229128
In a $500 \mathrm{~mL}$ flask, the degree of dissociation of $\mathrm{PCl}_{5}$ at equilibrium is $40 \%$ and the initial amount is 5 moles. The value of equilibrium constant in mol $\mathrm{L}^{-1}$ for the decomposition of $\mathrm{PCl}_{5}$ is
229078
Van't Hoff factor of $\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}$ is
1 One
2 Two
3 Three
4 Four
Explanation:
$\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2} \rightleftharpoons \mathrm{Ca}^{2+}+2 \mathrm{NO}_{3}^{-}$ It furnishes 3 ions per formula unit. So, its Van't Hoff factor is 3.
[J and K CET-(2013)]
Chemical Equilibrium
229095
$\mathrm{K}_{2} \mathrm{HgI}_{4}$ is $40 \%$ ionised in aqueous solution. The value of its Van't Hoff factor (i) is
1 1.6
2 1.8
3 2.2
4 2.0
Explanation:
The ionisation of $\mathrm{K}_{2} \mathrm{HgI}_{4}$ is given as$\mathrm{K}_{2} \mathrm{HgI}_{4} \longrightarrow 2 \mathrm{~K}^{+}+\mathrm{HgI}_{4}^{2-}$ $i=1+(\mathrm{n}-1) \alpha$ $\alpha=0.4, \mathrm{n}=3$ So, $\quad i=1+2 \times 0.4=1.8$
[JEE Mains 2019
Chemical Equilibrium
229121
If $\mathrm{BaCl}_{2}$ ionizes to an extent of $80 \%$ in aqueous solution, the value of van't Hoff factor is
1 2.6
2 0.4
3 0.8
4 2.4
Explanation:
Ionisation of $\mathrm{BaCl}_{2}$ is-
AP-EAMCET- (Engg.) - 2010
Chemical Equilibrium
229123
Van't Hoff factor of aq $\mathrm{K}_{2} \mathrm{SO}_{4}$ at infinite dilution has value equal to
1 1
2 2
3 3
4 between 2 and 3
Explanation:
At infinite dilution, complete dissociation will take place. So, $\mathrm{K}_{2} \mathrm{SO}_{4} \rightarrow 2 \mathrm{~K}^{+}+\mathrm{SO}_{4}{ }^{2-}$ van't Hoff factor $=2+1=3$
AMU-2009
Chemical Equilibrium
229128
In a $500 \mathrm{~mL}$ flask, the degree of dissociation of $\mathrm{PCl}_{5}$ at equilibrium is $40 \%$ and the initial amount is 5 moles. The value of equilibrium constant in mol $\mathrm{L}^{-1}$ for the decomposition of $\mathrm{PCl}_{5}$ is
