The resistance of $0.01 \ M$ $KCl$ solution at $298 \ K$ is $1500 \ \Omega$. If the conductivity of $0.01 \ M$ $KCl$ solution at $298 \ K$ is $0.1466 \times 10^{-3} \ S \ cm^{-1}$,the cell constant of the conductivity cell in $cm^{-1}$ is:

Which of the following electrolytic solutions has the least specific conductance (in $N$)?

$\Lambda_m^0(NH_4OH)$ is equal to:

Limiting molar conductivity of $NH_4OH$ [i.e.,$\Lambda ^o_{m(NH_4OH)}$] is equal to:

An aqueous solution of $X$ is added slowly to an aqueous solution of $Y$ as shown in List-$I$. The variation in conductivity of these reactions is given in List-$II$. Match List-$I$ with List-$II$ and select the correct answer using the code given below the lists :

List-$I$	List-$II$
$P$. $\underset{X}{(C_2H_5)_3N} + \underset{Y}{CH_3COOH}$	$1$. Conductivity decreases and then increases
$Q$. $\underset{X}{KI (0.1 \ M)} + \underset{Y}{AgNO_3 (0.01 \ M)}$	$2$. Conductivity decreases and then does not change much
$R$. $\underset{X}{CH_3COOH} + \underset{Y}{KOH}$	$3$. Conductivity increases and then does not change much
$S$. $\underset{X}{NaOH} + \underset{Y}{HI}$	$4$. Conductivity does not change much and then increases

Codes: $P \quad Q \quad R \quad S$

$A$ conductivity cell containing $5 \times 10^{-4} \ M \ NaCl$ solution develops a resistance of $14000 \ \Omega$ at $25^{\circ} C$. Calculate the conductivity of the solution if the cell constant is $0.84 \ cm^{-1}$.