Explanation

Correct answer is A

The graph is drawn using following equation,

${\mathrm{\Lambda }}_m={\mathrm{\Lambda }}_m^{\mathrm{\infty }}-b\sqrt{c}$

As the size of K⁺ is higher than the size of Na⁺, then the hydration radii of aqueous Na⁺ will be more than the aqueous K⁺. Therefore the ionic mobility of Na⁺ will be smaller than K⁺. Hence the conductance of K⁺ will be higher.

 $\therefore$ ${\mathrm{\Lambda }}_m^{\mathrm{\infty }}$ of K⁺ > ${\mathrm{\Lambda }}_m^{\mathrm{\infty }}$ of Na⁺

$\Rightarrow$ ${\mathrm{\Lambda }}_m^{\mathrm{\infty }}$(KCl) > ${\mathrm{\Lambda }}_m^{\mathrm{\infty }}$(NaCl)

So, y intercept is more for KCl.

As slope b is constant both the lines will be parallel.

Explanation

Correct answer is D

We know conductivity (k) = $\frac{G}{V}$

V = volume

When concentration decreases volume increases and when volume increases then conductivity (k) decreases.

So, we can say S1 is incorrect.

We know that,

${\lambda }_m=\frac{k}{c}$

where

$\lambda$_m = molar conductivity
k = conductivity
c = concentration

So, when concentration decreases molar conductivity increases.

So, we can say S2 is correct.

Explanation

Correct answer is B

${\wedge }_m^{\circ }$ (HA) = ${\wedge }_m^{\circ }$ (HCl) + ${\wedge }_m^{\circ }$ (NaA) $-$ ${\wedge }_m^{\circ }$ (NaCl)

= 425.9 + 100.5 $-$ 126.4

= 400 S cm² . mol^$-$1

    ${\wedge }_m^c$ = $\frac{K\times 1000}{M}$

= $\frac{5\times {10}^{-5}\times 1000}{{10}^{-3}}$

= 50 S cm² mol^$-$1

$\alpha$ = $\frac{{\mathrm{\Lambda }}_m^c}{{\mathrm{\Lambda }}_m^o}$ = $\frac{50}{400}$ = 0.125

Explanation

Correct answer is C

Given for $0.2$ $M$ solution

$R=50\mathrm{\Omega }$

$\kappa =1.45S{m}^{-1}=1.4\times {10}^{-2}Sc{m}^{-1}$

Now, $R=\rho \frac{\ell }{a}=\frac{1}{\kappa }\times \frac{\ell }{a}$

$\Rightarrow \frac{\ell }{a}=R\times \kappa =50\times 1.4\times {10}^{-2}$

For $0.5$ $M$ solution

$R=280\mathrm{\Omega };\kappa =?$

$\frac{\ell }{a}=50\times 1.4\times {10}^{-2}$

$\Rightarrow R=\rho \frac{\ell }{a}=\frac{1}{\kappa }\times \frac{\ell }{a}$

$\Rightarrow \kappa =\frac{1}{280}\times 50\times 1.4\times {10}^{-2}$

$=\frac{1}{280}\times 70\times {10}^{-2}=2.5\times {10}^{-3}Sc{m}^{-1}$

Now, ${\mathrm{\Lambda }}_m=\frac{\kappa \times 1000}{M}=\frac{2.5\times {10}^{-3}\times 1000}{0.5}$

$=5Sc{m}^{2}mo{l}^{-1}=5\times {10}^{-4}S{m}^{2}mo{l}^{-1}$

Explanation

Correct answer is C

According to Debye Huckle onsager equation,

${\lambda }_C={\lambda }_{\mathrm{\infty }}-B\sqrt{C}$

Correct Answer is Option (B)

NOTE : According to Kohlrausch's law, molar conductivity of weak electrolyte acetic acid $\left(C{H}_{3}COOH\right)$ can be calculated as follows:

${{\mathrm{\Lambda }}^o}_{C{H}_{3}COOH}=\left({{\mathrm{\Lambda }}^o}_{C{H}_{3}COONa}+{{\mathrm{\Lambda }}^o}_{HCl}\right)-{{\mathrm{\Lambda }}^o}_{NaCl}$

$\therefore$ Value of ${{\mathrm{\Lambda }}^o}_{NaCl}$ should also be known

for calculating value of ${{\mathrm{\Lambda }}^o}_{C{H}_{3}COOH}$

Correct Answer is Option (D)

${\mathrm{\Lambda }}_{C{H}_{3}COOH}^o$ is given by the following equation

${\mathrm{\Lambda }}_{C{H}_{3}COOH}^o=\left({\mathrm{\Lambda }}_{C{H}_{3}COONa}^o+{\mathrm{\Lambda }}_{HCl}^o\right)-\left({\mathrm{\Lambda }}_{NaCl}^o\right)$

Hence ${\mathrm{\Lambda }}_{NaCl}^{\circ }$ is required.

Correct Answer is Option (D)

$R=100\mathrm{\Omega },\kappa =\frac{1}{R}\left(\frac{l}{a}\right),\frac{l}{a}$ (cell constant)

$$ $$ $=1.29\times 100m^{-1}$

Given, $R=520\mathrm{\Omega },C=0.2M,\mu$ (molar conductivity) $=$ ?

$\mu =\kappa \times V$ ($\kappa$ can be calculated as $\kappa =\frac{1}{R}\left(\frac{1}{a}\right)$

now cell constant is known.

Hence, $\mu =\frac{1}{520}\times 129\times \frac{1000}{0.2}\times {10}^{-6}{m}^3$

$=12.4\times {10}^{-4}S{m}^{2}mo{l}^{-1}$

Correct Answer is Option (D)

Thus difluoro acetic acid being strongest acid will furnish maximum number of ions showing highest electrical conductivity. The decreasing acidic strength of the carboxylic acids given is difluoro acetic acid $>$ fluoro acetic acid $>$ chloro acitic acid $>$ acetic acid.

Correct Answer is Option (C)

$A_{HCl}^{\mathrm{\infty }}=426.2...\left(i\right)$

$A_{AcONa}^{\mathrm{\infty }}=91.0...\left(ii\right)$

$A_{NaCl}^{\mathrm{\infty }}=126.5...\left(iii\right)$

$A_{AcOH}^{\mathrm{\infty }}=\left(i\right)+\left(ii\right)-\left(iii\right)$

$=\left[426.2+91.0-126.5\right]$

$=390.7$

Correct Answer is Option (B)

Given $S\propto \frac{area\times conc}{\ell }=\frac{\kappa m^{2}mol}{m\times m^3}$

$\therefore$ $\kappa =S{m}^{2}mo{l}^{-1}$