Explanation

Correct answer is D

Ion	$H^{+}$	$K^{+}$	$C{l}^{-}$	$C{H}_{3}CO{O}^{-}$
${\mathrm{\Lambda }}_{mSc{m}^2/mole}^{\mathrm{\infty }}$	349.8	73.5	76.3	40.9

So,

${\mathrm{\Lambda }}_{mC{H}_{3}COOH}^{\mathrm{\infty }}={\mathrm{\Lambda }}_{m(H^{+})}^{\mathrm{\infty }}+{\mathrm{\Lambda }}_{mC{H}_{3}CO{O}^{-}}^{\mathrm{\infty }}$

= 349.8 + 40.9 = 390.7 Scm²/mole

${\mathrm{\Lambda }}_{mKCl}^{\mathrm{\infty }}={\mathrm{\Lambda }}_{m(K^{+})}^{\mathrm{\infty }}+{\mathrm{\Lambda }}_{m(C{l}^{-})}^{\mathrm{\infty }}$

= 73.5 + 76.3 = 149.3 Scm²/mole

So, Statement I is wrong or false.

As the concentration decreases, the dilution increases which increases the degree of dissociation, thus increasing the no. of ions, which increases the molar conductance.

So Statement II is false.

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 $=0.1\mathrm{M}$

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

correct answer is 26

# Explanation

Conductance (G) is reciprocal of resistance (R)

$R=\frac{1}{G}$ or $G=\frac{1}{R}=\frac{1}{0.243\mathrm{\Omega }}=4.115{\mathrm{\Omega }}^{-1}$

Relation between conductance (G),

conductivity ($\kappa$) and cell constant $\left(\frac{l}{A}\right)$ is given as

$\kappa =\frac{Gl}{A}$

$\Rightarrow \frac{l}{A}=\frac{\kappa }{G}=\frac{1.07\times {10}^{6}S{m}^{-1}}{4.115{\mathrm{\Omega }}^{-1}}=26\times {10}^4{m}^{-1}\Rightarrow x=26$

correct answer is 12

# Explanation

${\mathrm{\Lambda }}_m^{}=1000\times \frac{\kappa }{M}$

$=1000\times \frac{2\times {10}^{-5}}{0.001}=20$ S cm² mol^$-$1

$\Rightarrow \alpha =\frac{{\mathrm{\Lambda }}_m^{}}{{\mathrm{\Lambda }}_m^{\mathrm{\infty }}}=\frac{20}{190}=\left(\frac{2}{19}\right)$

HA $\rightleftharpoons$ H⁺ + A^$-$

$0.001(1-\alpha )0.001\alpha 0.001\alpha$

$\Rightarrow k_a=0.001\left(\frac{{\alpha }^2}{1-\alpha }\right)=\frac{0.001\times {\left(\frac{2}{19}\right)}^2}{1-\left(\frac{2}{19}\right)}$

$=12.3\times {10}^{-6}$

correct answer is 288

# Explanation

From Kohlrausch's law

${\mathrm{\Lambda }}_m^{\mathrm{\infty }}(BaS{O}_4)={\lambda }_m^{\mathrm{\infty }}(B{a}^{2+})+{\lambda }_m^{\mathrm{\infty }}(S{O}_4^{2-})$

${\mathrm{\Lambda }}_m^{\mathrm{\infty }}(BaS{O}_4)={\mathrm{\Lambda }}_m^{\mathrm{\infty }}(BaC{l}_2)+{\mathrm{\Lambda }}_m^{\mathrm{\infty }}(H_{2}S{O}_4)-2{\mathrm{\Lambda }}_m^{\mathrm{\infty }}(HCl)$

$=280+860-2(426)$

$=288$ S cm² mol^$-$1

correct answer is 57

# Explanation

For KCl solution,

$R=\left(\frac{1}{K}\right)\left(\frac{l}{A}\right)\Rightarrow \frac{l}{A}=R\times K=4.19\times 0.14$

= 0.58

For HCl solution,

$R=\left(\frac{1}{K}\right)\left(\frac{l}{A}\right)$

$\Rightarrow K=\frac{(l/A)}{R}=\frac{0.58}{1.03}=0.56=56\times {10}^{-2}$ Sm^$-$1

correct answer is 14

# Explanation

Conductance = $\frac{Conductivity}{Cellcons\tan t}$

$\therefore$ Conductivity = 0.55 $\times$ 10^$-$3 $\times$ 1.3 S cm^$-$1

Molar conductivity = $\frac{Conductivity(Sc{m}^{-1})\times 1000}{Molarity(mol/L)}$

$=\frac{0.55\times {10}^{-3}\times 1.3\times 100}{5\times {10}^{-3}}$

= 143 S cm² mol^$-$1

= 14.3 mS m² mol^$-$1

$\approx$ 14 mS m² mol^$-$1

Explanation

Correct answer is B

The electrolyte (X) must be weak electrolyte as such type of variation is always for weak electrolyte. So

X is CH₃COOH.

Explanation

Correct answer is D

BaSO₄ is sparingly soluble salt but NaCl is completely soluble salt so it will produce more number of ions. That is why
Conductance (NaCl) > Conductance (BaSO₄)

Explanation

Correct answer is B

Left hand side :
${\left({\mathrm{\Lambda }}_m^0\right)}_{NaBr}-{\left({\mathrm{\Lambda }}_m^0\right)}_{NaI}$ = ${\left({\mathrm{\Lambda }}_m^0\right)}_{B{r}^{-}}-{\left({\mathrm{\Lambda }}_m^0\right)}_{I^{-}}$ ....(1)

Right hand side :

${\left({\mathrm{\Lambda }}_m^0\right)}_{KBr}-{\left({\mathrm{\Lambda }}_m^0\right)}_{NaBr}$ = ${\left({\mathrm{\Lambda }}_m^0\right)}_{K^{+}}-{\left({\mathrm{\Lambda }}_m^0\right)}_{N{a}^{+}}$ ....(2)

According to Kohlrausch's law option (B) is incorrect as (1) and (2) are not equal.