correct answer is 3

# Explanation

Let us evaluate each statement one by one to determine whether it is true or false:

(A) Conductivity always decreases with decrease in concentration for both strong and weak electrolytes.

This statement is true. Conductivity is defined as the ability of a solution to conduct electricity, and it depends on the concentration of ions in the solution. As the concentration of ions decreases, the conductivity of the solution also decreases, regardless of whether the electrolyte is strong or weak.

(B) The number of ions per unit volume that carry current in a solution increases on dilution.

This statement is false. The number of ions per unit volume in a solution decreases on dilution, because the total number of ions in the solution remains the same, but the volume increases.

(C) Molar conductivity increases with decrease in concentration.

This statement is generally true. Molar conductivity is a measure of the ability of an electrolyte to conduct electricity, and it depends on both the concentration of the electrolyte and the mobility of the ions. As the concentration of the electrolyte decreases, the molar conductivity increases because the ions become more separated from each other, and the electrostatic interactions between them become weaker, allowing them to move more freely in the solution.

(D) The variation in molar conductivity is different for strong and weak electrolytes.

This statement is true. The variation in molar conductivity with concentration is different for strong and weak electrolytes. Strong electrolytes dissociate completely in solution, so their molar conductivity increases rapidly with dilution, while weak electrolytes dissociate only partially, so their molar conductivity increases more slowly with dilution.

(E) For weak electrolytes, the change in molar conductivity with dilution is due to decrease in degree of dissociation.

This statement is false. For weak electrolytes, the change in molar conductivity with dilution is due to an increase in the degree of dissociation, not a decrease.

Therefore, there are $\boxed{{\displaystyle 3}}$ correct statements in the given options, namely (A), (C), (D).

correct answer is 66

# Explanation

Given that the specific conductance, $k$, is $5\times {10}^{-5}Sc{m}^{-1}$ and the concentration, $C$, is $0.0025M$, we can find the molar conductivity, ${\lambda }_m$, as follows:

${\lambda }_m=\frac{k}{C}\times 1000=\frac{5\times {10}^{-5}\times {10}^3}{0.0025}=\frac{5\times {10}^{-2}}{2.5\times {10}^{-3}}=20Sc{m}^{2}mo{l}^{-1}$

Next, we find the degree of dissociation, $\alpha$, by dividing ${\lambda }_m$ by the limiting molar conductivity, ${\lambda }_m^o$:

$\alpha =\frac{20}{400}=\frac{1}{20}$

Finally, we use the formula for the dissociation constant of a weak acid, $K_a$:

$K_a=\frac{C{\alpha }^2}{1-\alpha }=\frac{0.0025\times \frac{1}{20}\times \frac{1}{20}}{\frac{19}{20}}=\frac{0.0025}{19\times 20}=6.6\times {10}^{-6}=66\times {10}^{-7}$

So, the correct answer is 66.

correct answer is 13039

# Explanation

$1\times {10}^{-5}\mathrm{M}{\mathrm{AgNO}}_3$ is added to $1\mathrm{L}$ of saturated solution of $\mathrm{AgBr}$. The conductivity of this solution at $298\mathrm{K}$ is _____________ $\times {10}^{-8}\mathrm{S}{\mathrm{m}}^{-1}$.

[Given : ${\mathrm{K}}_{\mathrm{SP}}(\mathrm{AgBr})=4.9\times {10}^{-13}$ at $298\mathrm{K}$

$\begin{array}{rl}& {\lambda }_{{\mathrm{Ag}}^{+}}^0=6\times {10}^{-3}\mathrm{S}{\mathrm{m}}^2{\mathrm{mol}}^{-1}\\ & {\lambda }_{{\mathrm{Br}}^{-}}^0=8\times {10}^{-3}\mathrm{S}{\mathrm{m}}^2{\mathrm{mol}}^{-1}\\ & {\lambda }_{{\mathrm{NO}}_3^{-}}^0=7\times {10}^{-3}\mathrm{S}{\mathrm{m}}^2{\mathrm{mol}}^{-1}]\end{array}$

correct answer is 25

# Explanation

$\begin{array}{rl}\text{ Molar conductivity }& =\frac{\mathrm{k}\times 1000}{\mathrm{C}}\\ \\ & =\frac{\frac{1}{5\times {10}^{-3}}\times 1000}{0.8}\\ \\ & =\frac{{10}^6}{4}=0.25\times {10}^6\\ \\ & =25\times {10}^4{\mathrm{\Omega }}^{-1}{\mathrm{cm}}^2{\mathrm{mol}}^{-1}\end{array}$

correct answer is 2

# Explanation

(A) ${\mathrm{\Lambda }}_{\mathrm{m}}^{\circ }$ for 'A' cannot be obtained by extra polation.

(C) At infinite dilution, value of degree of dissociation approaches one.

$\therefore \mathrm{A}$ and $\mathrm{C}$ are incorrect

correct answer is 1000

# Explanation

$\frac{l}{A}=129{\mathrm{m}}^{-1}$

$\mathrm{KCl}$ solution $1\Rightarrow 74.5\mathrm{ppm},{\mathrm{R}}_1=100\mathrm{\Omega }$

$\mathrm{KCl}$ solution $2\Rightarrow 149\mathrm{ppm},{\mathrm{R}}_2=50\mathrm{\Omega }$

$\begin{array}{rl}& \text{ Here, }\frac{pp{m}_1}{pp{m}_2}=\frac{M_1}{M_2}=(\frac{w_{1/M_0}}{V}\times \frac{V}{w_{2/M_0}})\\ & \frac{{\mathrm{\Lambda }}_1}{{\mathrm{\Lambda }}_2}=\frac{k_1\times \frac{1000}{M_1}}{k_2\times \frac{1000}{M_2}}\\ & =\frac{k_1}{k_2}\times \frac{M_1}{M_2}\\ & =\frac{50}{100}\times 2\\ & =\frac{{\mathrm{\Lambda }}_1}{{\mathrm{\Lambda }}_2}=1000\times {10}^{-3}\\ & =1000\end{array}$

correct answer is 14

# Explanation

Given

(1) ${\lambda }_m^{\mathrm{\infty }}(\mathrm{NaI})=12.7\mathrm{mS}{\mathrm{m}}^2{\mathrm{mol}}^{-1}$

(2) ${\lambda }_{\mathrm{m}}^{\mathrm{\infty }}\left({\mathrm{NaNO}}_3\right)=12.0\mathrm{mS}{\mathrm{m}}^2{\mathrm{mol}}^{-1}$

(3) ${\lambda }_{\mathrm{m}}^{\mathrm{\infty }}\left({\mathrm{AgNO}}_3\right)=13.3\mathrm{mS}{\mathrm{m}}^2{\mathrm{mol}}^{-1}$

${\lambda }_{\mathrm{m}}^{\mathrm{\infty }}(\mathrm{Ag}\mathrm{I})=(1)+(3)-(2)$

$=12.7+13.3-12.0$

$=26.0-12.0$

${\lambda }_{\mathrm{m}}^{\mathrm{\infty }}(\mathrm{Ag}\mathrm{I})=14.0$

correct answer is 266

# Explanation

Molarity of $\mathrm{KCl}$ solution $=0.1\mathrm{M}$

$\begin{array}{ll}\text{ Resistance }& =1750\mathrm{ohm}\\ \\ \text{ Conductivity }& =0.152\times {10}^{-3}\mathrm{S}{\mathrm{cm}}^{-1}\\ \\ \text{ Conductivity }& =\frac{\text{ Cell constant }}{\text{ Resistance }}\\ \\ \therefore \text{ Cell constant }& =0.152\times {10}^{-3}\times 1750\\ \\ & =266\times {10}^{-3}{\mathrm{cm}}^{-1}\end{array}$

Explanation

Correct answer is B

Statement I is false : KI is a strong electrolyte which gets completely dissociate on dissolution. On dilution, the molar conductivity almost remain constant.

Statement II is false : carbonic acid is a weak electrolyte which do not gets completely dissociate on dissolution. On dilution the molar conductivity of weak electrolyte increases sharply.



$\therefore$ Both Statements are false.

Explanation

Correct answer is A

${\mathrm{\Lambda }}_{{\mathrm{m}}_1}=\frac{{\mathrm{k}}_1\times 1000}{{\mathrm{M}}_1}=\frac{\mathrm{k}\times 1000}{\frac{10}{0.02}}$

${\mathrm{\Lambda }}_{{\mathrm{m}}_2}=\frac{{\mathrm{k}}_2\times 1000}{\frac{20}{0.08}}$

It is given that ${\mathrm{k}}_1={\mathrm{k}}_2$

${\mathrm{k}}_1=\frac{{\mathrm{\Lambda }}_{{\mathrm{m}}_1}}{2}{\mathrm{k}}_2=\frac{{\mathrm{\Lambda }}_{{\mathrm{m}}_2}}{4}$

Applying the given condition on conductivity.

$\begin{array}{c}\frac{{\mathrm{\Lambda }}_{{\mathrm{m}}_1}}{2}=\frac{{\mathrm{\Lambda }}_{{\mathrm{m}}_2}}{4}\\ {\mathrm{\Lambda }}_{{\mathrm{m}}_2}=2{\mathrm{\Lambda }}_{{\mathrm{m}}_1}\end{array}$

Explanation

Correct answer is A

Cell constant = $\left(\frac{l}{A}\right)$ $\Rightarrow$ Units = m^$-$1

Molar conductivity ($\mathrm{\Lambda }$_m) $\Rightarrow$ Units = Sm² mole^$-$1

Conductivity (K) $\Rightarrow$ Units = S m^$-$1

Degree of dissociation ($\alpha$ $\to$ Dimensionless

$\therefore$ (a) - (iii), (b) - (i), (c) - (iv), (d) - (ii)