Kohlrausch Law

Q. At $298K$, the limiting molar conductivity of a weak monobasic acid is $4\times {10}^{2}Sc{m}^{2}mo{l}^{-1}$. At $298K$, for an aqueous solution of the acid the degree of dissociation is $\alpha$ and the molar conductivity is $y\times {10}^{2}Sc{m}^{2}mo{l}^{-1}$ . At $298K$, upon $20$ times dilution with water, the molar conductivity of the solution becomes $3y\times {10}^{2}Sc{m}^{2}mo{l}^{-1}$

The value of $y$ is

Q. The difference in limiting molar conductivity between $KCl$ and $KN{O}_3$ is the same as the difference between :

1. $NaCl$ and $KN{O}_3$
2. $KCl$ and $NaCl$
3. $NaCl$ and $NaN{O}_3$
4. $KN{O}_3$ and $NaN{O}_3$

Q. The molar conductance at infinite dilution of $BaC{l}_2$, $NaCl$, $NaOH$ are
$280\times {10}^{-4}$, $126.5\times {10}^{-4}$, $248\times {10}^{-4}S{m}^{2}mo{l}^{-1}$ respectively. The molar conductance at infinite dilution for $Ba(OH{)}_2$ is:

1. $523\times {10}^{-4}S{m}^{2}mo{l}^{-1}$
2. $52.3\times {10}^{-4}S{m}^{2}mo{l}^{-1}$
3. $5.23\times {10}^{-4}S{m}^{2}mo{l}^{-1}$
4. $65\times {10}^{-4}S{m}^{2}mo{l}^{-1}$

Q. The limiting molar conductivities ${\lambda }^{\circ }$ for $NaCl,KBr$ and $KCl$ are $126,152$ and $150Sc{m}^{2}mo{l}^{-1}$respectively. The ${\lambda }^{\circ }$ for $NaBr$ is

1. $278Sc{m}^{2}mo{l}^{-1}$
2. $176Sc{m}^{2}mo{l}^{-1}$
3. $128Sc{m}^{2}mo{l}^{-1}$
4. $302Sc{m}^{2}mo{l}^{-}1$

Q. Degree of dissociation $\alpha$ =

1. $\frac{{\wedge }_m}{{\wedge }_m^{\circ }}$
2. $\frac{{\wedge }_m^{\circ }}{{\wedge }_m}$
3. ${\wedge }_m-{\wedge }_m^{\circ }$
4. $\sqrt{{\wedge }_m}-\sqrt{{\wedge }_m^{\circ }}$

Q. How do density varies with dilution

Q. The conductance of a 0.0015 M aqueous solution of a weak monobasic acid was determined by using a conductivity cell consisting of platinized Pt electrodes. The distance between the electrodes is 120 cm with an area of cross section of $1c{m}^2$. The conductance of this solution was found to be $5\times {10}^{-7}S$. The pH of the solution is 4. The value of limiting molar conductivity $\left({\Lambda }_m^{\circ }\right)$ of this weak monobasic acid in aqueous solution is $Z\times {10}^{2}Sc{m}^{-1}mo{l}^{-1}$. The value of Z is___

Q. ${\wedge }_m^{\circ }\left(N{H}_{4}OH\right)$ is equal to

1. ${\wedge }_m^{\circ }\left(N{H}_{4}OH\right)+{\wedge }_m^{\circ }\left(N{H}_{4}Cl\right)-{\wedge }_m^{\circ }\left(HCl\right)$
2. ${\wedge }_m^{\circ }\left(N{H}_{4}Cl\right)+{\wedge }_m^{\circ }\left(NaOH\right)-{\wedge }_m^{\circ }\left(NaCl\right)$
3. ${\wedge }_m^{\circ }\left(N{H}_{4}Cl\right)+{\wedge }_m^{\circ }\left(NaCl\right)-{\wedge }_m^{\circ }\left(NaOH\right)$
4. ${\wedge }_m^{\circ }\left(NaOH\right)+{\wedge }_m^{\circ }\left(NaCl\right)-{\wedge }_m^{\circ }\left(N{H}_{4}Cl\right)$

Q. According to Kohlrausch’s law, conductivity of ions at infinite dilutions does not depend on

1. nature of co-ions
2. temperature
3. pressure
4. concentration

Q. According to Kohlrausch’s law, conductivity of ions is constant at

1. fixed temperature
2. infinite dilution
3. 1 M concentration
4. all the concentrations

Q. Which of the following options are correct for solubility and solubility product of
$C{O}_2\left[Fe{\left(CN\right)}_6\right]$ in water at ${25}^{\circ }$. Conductivity of a saturated solution of $C{O}_2\left[Fe{\left(CN\right)}_6\right]$ is $2.06\times {10}^{-6}{\Omega }^{-1}{\text{cm}}^{-1}$ and that of water used is $4.1\times {10}^{-7}{\Omega }^{-1}{\text{cm}}^{-1}.$
The ionic molar conductivities of $C{o}^{2+}\text{and}Fe(CN{)}_6^{4-}$ are $86.0{\Omega }^{-1}{\text{cm}}^2{\text{mol}}^{-1}\text{and}444.0{\Omega }^{-1}{\text{cm}}^2{\text{mol}}^{-1},$ respectively.

1. $K_{sp}=7.69\times {10}^{-17}$
2. $K_{sp}=7.6872\times {10}^{-14}$
3. $\text{Solubility =}2.6\times {10}^{-6}\text{mol/L}$
4. $\text{Solubility =}2.6\times {10}^{-5}\text{mol/L}$

Q. Which of the following graph is correct?

${\Lambda }_m$ - Molar conductivity
$C$ - concentration of electrolyte

1. 
2. 
3. 
4.