229467
The $K_H$ value (K bar) of Argon (I). carbondioxide (II), formaldehyde (III) and methane (IV) are respectively $40.3,167,1.83$ $\times 10^{-5}$ and 0.413 at $298 \mathrm{~K}$. The increasing order of solubility of gas in liquid is
1 I $<$ II $<$ IV $<$ III
2 III $<$ IV $<$ II $<$ I
3 I $<$ III $<$ II $<$ IV
4 I $<$ IV $<$ II $<$ III
Explanation:
Higher the value of the $\mathrm{K}_{\mathrm{H}}$, the lower is the solubility of gas in liquid. Hence, the order of increasing solubility will be- $\mathrm{Ar}<\mathrm{CO}_2<\mathrm{CH}_4<\mathrm{HCHO}$
Karnataka-CET-2021
Ionic Equilibrium
229477
Which among the following salts, solubility decreases with increase in temperature?
1 $\mathrm{Na}_2 \mathrm{SO}_4$
2 $\mathrm{NaBr}$
3 $\mathrm{NaCl}$
4 $\mathrm{KCl}$
Explanation:
For $\mathrm{Na}_2 \mathrm{SO}_4$ salt, Solubility decreases with Increase in temperature because reaction of $\mathrm{Na}_2 \mathrm{SO}_4$ with water is an exothermic reaction. As a result solubility decreases. For $\mathrm{NaBr}, \mathrm{NaCl}, \mathrm{KCl}$ the dissolution process i.e. endothermic
Shift-I
Ionic Equilibrium
229444
The correct order of solubility of the sulphates of alkaline earth metals in water is
Down the group the solutility of sulphates decreases So the sulphates of $\mathrm{Be}$ is more soluble and the sulphates of $\mathrm{Ba}$ is least soluble. So, the order of solubility of Sulfate is- $\mathrm{Be}>\mathrm{Mg}>\mathrm{Ca}>\mathrm{Sr}>\mathrm{Ba}$
J and K CET-(2004)
Ionic Equilibrium
229357
The solubility of AgCl will be maximum in which of the following?
1 $0.01 \mathrm{M} \mathrm{KCl}$
2 $0.01 \mathrm{M} \mathrm{HCl}$
3 $0.01 \mathrm{M} \mathrm{AgNO}_3$
4 Deionized water
Explanation:
If similar ion is present already in solution, then the solubility of the salt decreases due to common ion effect. So we need to dissolve $\mathrm{AgCl}$ having $\mathrm{Ag}^{+}$and $\mathrm{Cl}^{-}$ions. So maximum solubility of $\mathrm{AgCl}$ will be in water.
229467
The $K_H$ value (K bar) of Argon (I). carbondioxide (II), formaldehyde (III) and methane (IV) are respectively $40.3,167,1.83$ $\times 10^{-5}$ and 0.413 at $298 \mathrm{~K}$. The increasing order of solubility of gas in liquid is
1 I $<$ II $<$ IV $<$ III
2 III $<$ IV $<$ II $<$ I
3 I $<$ III $<$ II $<$ IV
4 I $<$ IV $<$ II $<$ III
Explanation:
Higher the value of the $\mathrm{K}_{\mathrm{H}}$, the lower is the solubility of gas in liquid. Hence, the order of increasing solubility will be- $\mathrm{Ar}<\mathrm{CO}_2<\mathrm{CH}_4<\mathrm{HCHO}$
Karnataka-CET-2021
Ionic Equilibrium
229477
Which among the following salts, solubility decreases with increase in temperature?
1 $\mathrm{Na}_2 \mathrm{SO}_4$
2 $\mathrm{NaBr}$
3 $\mathrm{NaCl}$
4 $\mathrm{KCl}$
Explanation:
For $\mathrm{Na}_2 \mathrm{SO}_4$ salt, Solubility decreases with Increase in temperature because reaction of $\mathrm{Na}_2 \mathrm{SO}_4$ with water is an exothermic reaction. As a result solubility decreases. For $\mathrm{NaBr}, \mathrm{NaCl}, \mathrm{KCl}$ the dissolution process i.e. endothermic
Shift-I
Ionic Equilibrium
229444
The correct order of solubility of the sulphates of alkaline earth metals in water is
Down the group the solutility of sulphates decreases So the sulphates of $\mathrm{Be}$ is more soluble and the sulphates of $\mathrm{Ba}$ is least soluble. So, the order of solubility of Sulfate is- $\mathrm{Be}>\mathrm{Mg}>\mathrm{Ca}>\mathrm{Sr}>\mathrm{Ba}$
J and K CET-(2004)
Ionic Equilibrium
229357
The solubility of AgCl will be maximum in which of the following?
1 $0.01 \mathrm{M} \mathrm{KCl}$
2 $0.01 \mathrm{M} \mathrm{HCl}$
3 $0.01 \mathrm{M} \mathrm{AgNO}_3$
4 Deionized water
Explanation:
If similar ion is present already in solution, then the solubility of the salt decreases due to common ion effect. So we need to dissolve $\mathrm{AgCl}$ having $\mathrm{Ag}^{+}$and $\mathrm{Cl}^{-}$ions. So maximum solubility of $\mathrm{AgCl}$ will be in water.
229467
The $K_H$ value (K bar) of Argon (I). carbondioxide (II), formaldehyde (III) and methane (IV) are respectively $40.3,167,1.83$ $\times 10^{-5}$ and 0.413 at $298 \mathrm{~K}$. The increasing order of solubility of gas in liquid is
1 I $<$ II $<$ IV $<$ III
2 III $<$ IV $<$ II $<$ I
3 I $<$ III $<$ II $<$ IV
4 I $<$ IV $<$ II $<$ III
Explanation:
Higher the value of the $\mathrm{K}_{\mathrm{H}}$, the lower is the solubility of gas in liquid. Hence, the order of increasing solubility will be- $\mathrm{Ar}<\mathrm{CO}_2<\mathrm{CH}_4<\mathrm{HCHO}$
Karnataka-CET-2021
Ionic Equilibrium
229477
Which among the following salts, solubility decreases with increase in temperature?
1 $\mathrm{Na}_2 \mathrm{SO}_4$
2 $\mathrm{NaBr}$
3 $\mathrm{NaCl}$
4 $\mathrm{KCl}$
Explanation:
For $\mathrm{Na}_2 \mathrm{SO}_4$ salt, Solubility decreases with Increase in temperature because reaction of $\mathrm{Na}_2 \mathrm{SO}_4$ with water is an exothermic reaction. As a result solubility decreases. For $\mathrm{NaBr}, \mathrm{NaCl}, \mathrm{KCl}$ the dissolution process i.e. endothermic
Shift-I
Ionic Equilibrium
229444
The correct order of solubility of the sulphates of alkaline earth metals in water is
Down the group the solutility of sulphates decreases So the sulphates of $\mathrm{Be}$ is more soluble and the sulphates of $\mathrm{Ba}$ is least soluble. So, the order of solubility of Sulfate is- $\mathrm{Be}>\mathrm{Mg}>\mathrm{Ca}>\mathrm{Sr}>\mathrm{Ba}$
J and K CET-(2004)
Ionic Equilibrium
229357
The solubility of AgCl will be maximum in which of the following?
1 $0.01 \mathrm{M} \mathrm{KCl}$
2 $0.01 \mathrm{M} \mathrm{HCl}$
3 $0.01 \mathrm{M} \mathrm{AgNO}_3$
4 Deionized water
Explanation:
If similar ion is present already in solution, then the solubility of the salt decreases due to common ion effect. So we need to dissolve $\mathrm{AgCl}$ having $\mathrm{Ag}^{+}$and $\mathrm{Cl}^{-}$ions. So maximum solubility of $\mathrm{AgCl}$ will be in water.
229467
The $K_H$ value (K bar) of Argon (I). carbondioxide (II), formaldehyde (III) and methane (IV) are respectively $40.3,167,1.83$ $\times 10^{-5}$ and 0.413 at $298 \mathrm{~K}$. The increasing order of solubility of gas in liquid is
1 I $<$ II $<$ IV $<$ III
2 III $<$ IV $<$ II $<$ I
3 I $<$ III $<$ II $<$ IV
4 I $<$ IV $<$ II $<$ III
Explanation:
Higher the value of the $\mathrm{K}_{\mathrm{H}}$, the lower is the solubility of gas in liquid. Hence, the order of increasing solubility will be- $\mathrm{Ar}<\mathrm{CO}_2<\mathrm{CH}_4<\mathrm{HCHO}$
Karnataka-CET-2021
Ionic Equilibrium
229477
Which among the following salts, solubility decreases with increase in temperature?
1 $\mathrm{Na}_2 \mathrm{SO}_4$
2 $\mathrm{NaBr}$
3 $\mathrm{NaCl}$
4 $\mathrm{KCl}$
Explanation:
For $\mathrm{Na}_2 \mathrm{SO}_4$ salt, Solubility decreases with Increase in temperature because reaction of $\mathrm{Na}_2 \mathrm{SO}_4$ with water is an exothermic reaction. As a result solubility decreases. For $\mathrm{NaBr}, \mathrm{NaCl}, \mathrm{KCl}$ the dissolution process i.e. endothermic
Shift-I
Ionic Equilibrium
229444
The correct order of solubility of the sulphates of alkaline earth metals in water is
Down the group the solutility of sulphates decreases So the sulphates of $\mathrm{Be}$ is more soluble and the sulphates of $\mathrm{Ba}$ is least soluble. So, the order of solubility of Sulfate is- $\mathrm{Be}>\mathrm{Mg}>\mathrm{Ca}>\mathrm{Sr}>\mathrm{Ba}$
J and K CET-(2004)
Ionic Equilibrium
229357
The solubility of AgCl will be maximum in which of the following?
1 $0.01 \mathrm{M} \mathrm{KCl}$
2 $0.01 \mathrm{M} \mathrm{HCl}$
3 $0.01 \mathrm{M} \mathrm{AgNO}_3$
4 Deionized water
Explanation:
If similar ion is present already in solution, then the solubility of the salt decreases due to common ion effect. So we need to dissolve $\mathrm{AgCl}$ having $\mathrm{Ag}^{+}$and $\mathrm{Cl}^{-}$ions. So maximum solubility of $\mathrm{AgCl}$ will be in water.
