229402
The solubility product of $\mathrm{AgCl}$ is $4.0 \mathrm{10}^{-10}$ at $298 \mathrm{~K}$. The solubility of $\mathrm{AgCl}$ in $0.04 \mathrm{M} \mathrm{CaCl}_2$ will be
229403
The solubility of $\mathrm{AgCl}$ is $1 \times \mathbf{1 0}^{-5} \mathrm{~mol} /$ litre. Its solubility in 0.1 molar sodium chloride solution is
Let solubility of $\mathrm{M}_2 \mathrm{X}$ is $\mathrm{S}_1, M X$ is $S_2$ and $\mathrm{MX}_3$ is $\mathrm{S}_3$ $\mathrm{S}_1=\left(\mathrm{K}_{\mathrm{sp}} / 4\right)^{-\frac{1}{3}} ; \mathrm{S}_2=\left(\mathrm{K}_{\mathrm{sp}}\right)^{1 / 2}$ $\mathrm{S}_3=\left(\mathrm{K}_{\mathrm{sp}} / 27\right)^{1 / 4}$ As $\mathrm{K}_{\mathrm{sp}}$ Let solubility of $10^{-12}$ (say) in that case $\mathrm{S}_1=\frac{1}{3} \sqrt{4}, \mathrm{~S}_2=10^6 \text { and } \mathrm{S}_3=\frac{1}{3 \sqrt{3}} 10^{-3}$ The correct order of solubilities is $\mathrm{MX}_3>\mathrm{M}_2 \mathrm{X}>\mathrm{MX}$
Assam CEE-2014
Ionic Equilibrium
229405
If $\mathrm{S}$ is the solubility of $\mathrm{Zr}_3\left(\mathrm{PO}_4\right)_4$ in Pure water then.
229402
The solubility product of $\mathrm{AgCl}$ is $4.0 \mathrm{10}^{-10}$ at $298 \mathrm{~K}$. The solubility of $\mathrm{AgCl}$ in $0.04 \mathrm{M} \mathrm{CaCl}_2$ will be
229403
The solubility of $\mathrm{AgCl}$ is $1 \times \mathbf{1 0}^{-5} \mathrm{~mol} /$ litre. Its solubility in 0.1 molar sodium chloride solution is
Let solubility of $\mathrm{M}_2 \mathrm{X}$ is $\mathrm{S}_1, M X$ is $S_2$ and $\mathrm{MX}_3$ is $\mathrm{S}_3$ $\mathrm{S}_1=\left(\mathrm{K}_{\mathrm{sp}} / 4\right)^{-\frac{1}{3}} ; \mathrm{S}_2=\left(\mathrm{K}_{\mathrm{sp}}\right)^{1 / 2}$ $\mathrm{S}_3=\left(\mathrm{K}_{\mathrm{sp}} / 27\right)^{1 / 4}$ As $\mathrm{K}_{\mathrm{sp}}$ Let solubility of $10^{-12}$ (say) in that case $\mathrm{S}_1=\frac{1}{3} \sqrt{4}, \mathrm{~S}_2=10^6 \text { and } \mathrm{S}_3=\frac{1}{3 \sqrt{3}} 10^{-3}$ The correct order of solubilities is $\mathrm{MX}_3>\mathrm{M}_2 \mathrm{X}>\mathrm{MX}$
Assam CEE-2014
Ionic Equilibrium
229405
If $\mathrm{S}$ is the solubility of $\mathrm{Zr}_3\left(\mathrm{PO}_4\right)_4$ in Pure water then.
229402
The solubility product of $\mathrm{AgCl}$ is $4.0 \mathrm{10}^{-10}$ at $298 \mathrm{~K}$. The solubility of $\mathrm{AgCl}$ in $0.04 \mathrm{M} \mathrm{CaCl}_2$ will be
229403
The solubility of $\mathrm{AgCl}$ is $1 \times \mathbf{1 0}^{-5} \mathrm{~mol} /$ litre. Its solubility in 0.1 molar sodium chloride solution is
Let solubility of $\mathrm{M}_2 \mathrm{X}$ is $\mathrm{S}_1, M X$ is $S_2$ and $\mathrm{MX}_3$ is $\mathrm{S}_3$ $\mathrm{S}_1=\left(\mathrm{K}_{\mathrm{sp}} / 4\right)^{-\frac{1}{3}} ; \mathrm{S}_2=\left(\mathrm{K}_{\mathrm{sp}}\right)^{1 / 2}$ $\mathrm{S}_3=\left(\mathrm{K}_{\mathrm{sp}} / 27\right)^{1 / 4}$ As $\mathrm{K}_{\mathrm{sp}}$ Let solubility of $10^{-12}$ (say) in that case $\mathrm{S}_1=\frac{1}{3} \sqrt{4}, \mathrm{~S}_2=10^6 \text { and } \mathrm{S}_3=\frac{1}{3 \sqrt{3}} 10^{-3}$ The correct order of solubilities is $\mathrm{MX}_3>\mathrm{M}_2 \mathrm{X}>\mathrm{MX}$
Assam CEE-2014
Ionic Equilibrium
229405
If $\mathrm{S}$ is the solubility of $\mathrm{Zr}_3\left(\mathrm{PO}_4\right)_4$ in Pure water then.
NEET Test Series from KOTA - 10 Papers In MS WORD
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Ionic Equilibrium
229402
The solubility product of $\mathrm{AgCl}$ is $4.0 \mathrm{10}^{-10}$ at $298 \mathrm{~K}$. The solubility of $\mathrm{AgCl}$ in $0.04 \mathrm{M} \mathrm{CaCl}_2$ will be
229403
The solubility of $\mathrm{AgCl}$ is $1 \times \mathbf{1 0}^{-5} \mathrm{~mol} /$ litre. Its solubility in 0.1 molar sodium chloride solution is
Let solubility of $\mathrm{M}_2 \mathrm{X}$ is $\mathrm{S}_1, M X$ is $S_2$ and $\mathrm{MX}_3$ is $\mathrm{S}_3$ $\mathrm{S}_1=\left(\mathrm{K}_{\mathrm{sp}} / 4\right)^{-\frac{1}{3}} ; \mathrm{S}_2=\left(\mathrm{K}_{\mathrm{sp}}\right)^{1 / 2}$ $\mathrm{S}_3=\left(\mathrm{K}_{\mathrm{sp}} / 27\right)^{1 / 4}$ As $\mathrm{K}_{\mathrm{sp}}$ Let solubility of $10^{-12}$ (say) in that case $\mathrm{S}_1=\frac{1}{3} \sqrt{4}, \mathrm{~S}_2=10^6 \text { and } \mathrm{S}_3=\frac{1}{3 \sqrt{3}} 10^{-3}$ The correct order of solubilities is $\mathrm{MX}_3>\mathrm{M}_2 \mathrm{X}>\mathrm{MX}$
Assam CEE-2014
Ionic Equilibrium
229405
If $\mathrm{S}$ is the solubility of $\mathrm{Zr}_3\left(\mathrm{PO}_4\right)_4$ in Pure water then.