$\mathrm{AgCl}$ is soluble in $\mathrm{NH}_{4} \mathrm{OH}$. The solubility is due to formation of $\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right]^{+}$. Silver chloride $(\mathrm{AgCl})$ is a white crystalline solid. $\mathrm{AgCl}$ reacts with ammonium hydroxide and yields diamine. $\underset{\text { Whiteppt }}{\mathrm{AgCl}(\mathrm{s})}+2 \mathrm{NH}_{4} \mathrm{OH}(\mathrm{aq}) \rightarrow \underset{\text { Solublecomplete }}{\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right] \mathrm{Cl}}$
JCECE - 2019
SOLUTIONS
277505
Which of the following compound is least soluble?
1 $\mathrm{Na}_{2} \mathrm{~S}$
2 $\mathrm{MgS}$
3 $\mathrm{MgCl}_{2}$
4 $\mathrm{NaCl}$
Explanation:
It is clear that higher the lattice energy lower the solubility. $\mathrm{MgS}$ is a bi-valent ionic solid and It has higher lattice energy.
AIIMS 26 May 2019 (Morning)
SOLUTIONS
277507
Which of the following conditions are correct for real solutions showing negative deviation from Raoult's law?
Conditions that are real solution showing negative deviation from Raoult's law are - $\mathrm{P}_{\mathrm{A}}<\mathrm{P}_{\mathrm{A}}^{\circ} \mathrm{X}_{\mathrm{A}}$ $\mathrm{P}_{\mathrm{B}}<\mathrm{P}_{\mathrm{B}}^{\circ} \mathrm{X}_{\mathrm{B}}$ So, $\Delta_{\mathrm{Mix}} \mathrm{H}=-\mathrm{ve}$ and dissolution is exothermic heating decreases solubility $\Delta_{\text {Mixing }} \mathrm{V}=-\mathrm{ve}$.
TS EAMCET-2017
SOLUTIONS
277510
$\mathrm{PbCI}_{2}$ is insoluble in cold water. Addition of HCI increases its solubility due to
1 formation of soluble complex anions like $\left[\mathrm{PbCl}_{3}\right]^{-}$
2 oxidation of $\mathrm{Pb}$ (II) to $\mathrm{PB}$ (IV)
3 formation of $\left[\mathrm{Pb}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$
4 formation of polymeric lead complex
Explanation:
Required Chemical Reaction, $\mathrm{PbCl}_{2}+\mathrm{Cl}^{-} \stackrel{\text { Cold }}{\longrightarrow}\left[\mathrm{PbCl}_{3}\right]^{-}$ $\mathrm{PbCl}_{2}+2 \mathrm{Cl}^{-} \stackrel{\text { increase }}{\longrightarrow}\left[\mathrm{PbCl}_{4}\right]^{2-}$ From above reaction it is clear that addition of $\mathrm{HCl}$ increases it's solubility due to form of soluble complex anion like $\left[\mathrm{PbCl}_{3}\right]^{-}$.
$\mathrm{AgCl}$ is soluble in $\mathrm{NH}_{4} \mathrm{OH}$. The solubility is due to formation of $\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right]^{+}$. Silver chloride $(\mathrm{AgCl})$ is a white crystalline solid. $\mathrm{AgCl}$ reacts with ammonium hydroxide and yields diamine. $\underset{\text { Whiteppt }}{\mathrm{AgCl}(\mathrm{s})}+2 \mathrm{NH}_{4} \mathrm{OH}(\mathrm{aq}) \rightarrow \underset{\text { Solublecomplete }}{\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right] \mathrm{Cl}}$
JCECE - 2019
SOLUTIONS
277505
Which of the following compound is least soluble?
1 $\mathrm{Na}_{2} \mathrm{~S}$
2 $\mathrm{MgS}$
3 $\mathrm{MgCl}_{2}$
4 $\mathrm{NaCl}$
Explanation:
It is clear that higher the lattice energy lower the solubility. $\mathrm{MgS}$ is a bi-valent ionic solid and It has higher lattice energy.
AIIMS 26 May 2019 (Morning)
SOLUTIONS
277507
Which of the following conditions are correct for real solutions showing negative deviation from Raoult's law?
Conditions that are real solution showing negative deviation from Raoult's law are - $\mathrm{P}_{\mathrm{A}}<\mathrm{P}_{\mathrm{A}}^{\circ} \mathrm{X}_{\mathrm{A}}$ $\mathrm{P}_{\mathrm{B}}<\mathrm{P}_{\mathrm{B}}^{\circ} \mathrm{X}_{\mathrm{B}}$ So, $\Delta_{\mathrm{Mix}} \mathrm{H}=-\mathrm{ve}$ and dissolution is exothermic heating decreases solubility $\Delta_{\text {Mixing }} \mathrm{V}=-\mathrm{ve}$.
TS EAMCET-2017
SOLUTIONS
277510
$\mathrm{PbCI}_{2}$ is insoluble in cold water. Addition of HCI increases its solubility due to
1 formation of soluble complex anions like $\left[\mathrm{PbCl}_{3}\right]^{-}$
2 oxidation of $\mathrm{Pb}$ (II) to $\mathrm{PB}$ (IV)
3 formation of $\left[\mathrm{Pb}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$
4 formation of polymeric lead complex
Explanation:
Required Chemical Reaction, $\mathrm{PbCl}_{2}+\mathrm{Cl}^{-} \stackrel{\text { Cold }}{\longrightarrow}\left[\mathrm{PbCl}_{3}\right]^{-}$ $\mathrm{PbCl}_{2}+2 \mathrm{Cl}^{-} \stackrel{\text { increase }}{\longrightarrow}\left[\mathrm{PbCl}_{4}\right]^{2-}$ From above reaction it is clear that addition of $\mathrm{HCl}$ increases it's solubility due to form of soluble complex anion like $\left[\mathrm{PbCl}_{3}\right]^{-}$.
$\mathrm{AgCl}$ is soluble in $\mathrm{NH}_{4} \mathrm{OH}$. The solubility is due to formation of $\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right]^{+}$. Silver chloride $(\mathrm{AgCl})$ is a white crystalline solid. $\mathrm{AgCl}$ reacts with ammonium hydroxide and yields diamine. $\underset{\text { Whiteppt }}{\mathrm{AgCl}(\mathrm{s})}+2 \mathrm{NH}_{4} \mathrm{OH}(\mathrm{aq}) \rightarrow \underset{\text { Solublecomplete }}{\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right] \mathrm{Cl}}$
JCECE - 2019
SOLUTIONS
277505
Which of the following compound is least soluble?
1 $\mathrm{Na}_{2} \mathrm{~S}$
2 $\mathrm{MgS}$
3 $\mathrm{MgCl}_{2}$
4 $\mathrm{NaCl}$
Explanation:
It is clear that higher the lattice energy lower the solubility. $\mathrm{MgS}$ is a bi-valent ionic solid and It has higher lattice energy.
AIIMS 26 May 2019 (Morning)
SOLUTIONS
277507
Which of the following conditions are correct for real solutions showing negative deviation from Raoult's law?
Conditions that are real solution showing negative deviation from Raoult's law are - $\mathrm{P}_{\mathrm{A}}<\mathrm{P}_{\mathrm{A}}^{\circ} \mathrm{X}_{\mathrm{A}}$ $\mathrm{P}_{\mathrm{B}}<\mathrm{P}_{\mathrm{B}}^{\circ} \mathrm{X}_{\mathrm{B}}$ So, $\Delta_{\mathrm{Mix}} \mathrm{H}=-\mathrm{ve}$ and dissolution is exothermic heating decreases solubility $\Delta_{\text {Mixing }} \mathrm{V}=-\mathrm{ve}$.
TS EAMCET-2017
SOLUTIONS
277510
$\mathrm{PbCI}_{2}$ is insoluble in cold water. Addition of HCI increases its solubility due to
1 formation of soluble complex anions like $\left[\mathrm{PbCl}_{3}\right]^{-}$
2 oxidation of $\mathrm{Pb}$ (II) to $\mathrm{PB}$ (IV)
3 formation of $\left[\mathrm{Pb}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$
4 formation of polymeric lead complex
Explanation:
Required Chemical Reaction, $\mathrm{PbCl}_{2}+\mathrm{Cl}^{-} \stackrel{\text { Cold }}{\longrightarrow}\left[\mathrm{PbCl}_{3}\right]^{-}$ $\mathrm{PbCl}_{2}+2 \mathrm{Cl}^{-} \stackrel{\text { increase }}{\longrightarrow}\left[\mathrm{PbCl}_{4}\right]^{2-}$ From above reaction it is clear that addition of $\mathrm{HCl}$ increases it's solubility due to form of soluble complex anion like $\left[\mathrm{PbCl}_{3}\right]^{-}$.
$\mathrm{AgCl}$ is soluble in $\mathrm{NH}_{4} \mathrm{OH}$. The solubility is due to formation of $\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right]^{+}$. Silver chloride $(\mathrm{AgCl})$ is a white crystalline solid. $\mathrm{AgCl}$ reacts with ammonium hydroxide and yields diamine. $\underset{\text { Whiteppt }}{\mathrm{AgCl}(\mathrm{s})}+2 \mathrm{NH}_{4} \mathrm{OH}(\mathrm{aq}) \rightarrow \underset{\text { Solublecomplete }}{\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right] \mathrm{Cl}}$
JCECE - 2019
SOLUTIONS
277505
Which of the following compound is least soluble?
1 $\mathrm{Na}_{2} \mathrm{~S}$
2 $\mathrm{MgS}$
3 $\mathrm{MgCl}_{2}$
4 $\mathrm{NaCl}$
Explanation:
It is clear that higher the lattice energy lower the solubility. $\mathrm{MgS}$ is a bi-valent ionic solid and It has higher lattice energy.
AIIMS 26 May 2019 (Morning)
SOLUTIONS
277507
Which of the following conditions are correct for real solutions showing negative deviation from Raoult's law?
Conditions that are real solution showing negative deviation from Raoult's law are - $\mathrm{P}_{\mathrm{A}}<\mathrm{P}_{\mathrm{A}}^{\circ} \mathrm{X}_{\mathrm{A}}$ $\mathrm{P}_{\mathrm{B}}<\mathrm{P}_{\mathrm{B}}^{\circ} \mathrm{X}_{\mathrm{B}}$ So, $\Delta_{\mathrm{Mix}} \mathrm{H}=-\mathrm{ve}$ and dissolution is exothermic heating decreases solubility $\Delta_{\text {Mixing }} \mathrm{V}=-\mathrm{ve}$.
TS EAMCET-2017
SOLUTIONS
277510
$\mathrm{PbCI}_{2}$ is insoluble in cold water. Addition of HCI increases its solubility due to
1 formation of soluble complex anions like $\left[\mathrm{PbCl}_{3}\right]^{-}$
2 oxidation of $\mathrm{Pb}$ (II) to $\mathrm{PB}$ (IV)
3 formation of $\left[\mathrm{Pb}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$
4 formation of polymeric lead complex
Explanation:
Required Chemical Reaction, $\mathrm{PbCl}_{2}+\mathrm{Cl}^{-} \stackrel{\text { Cold }}{\longrightarrow}\left[\mathrm{PbCl}_{3}\right]^{-}$ $\mathrm{PbCl}_{2}+2 \mathrm{Cl}^{-} \stackrel{\text { increase }}{\longrightarrow}\left[\mathrm{PbCl}_{4}\right]^{2-}$ From above reaction it is clear that addition of $\mathrm{HCl}$ increases it's solubility due to form of soluble complex anion like $\left[\mathrm{PbCl}_{3}\right]^{-}$.
