The resistance of a 0-1 N solution of a salt is found to be 2-5 - 103 ohm- The equivalent conductance of the solution is- Cell constant - 1-15 cm-1

The resistance of 0.1 N solution of a salt is found to be $2.5 \times 10^{3}$ . The equivalent conductance of the solution is: (cell constant=1.15 $c m^{- 1}$ )

Q. The resistance of

N / 10

solution is found to be

2.5 \times 10^{3} o h m

. The equivalent conductance of the solution is (cell constant

= 1.25 c m^{- 1})

Q. The resistance of

1 N

solution of

C H_{3} C O O H

250 o h m

when measured in a cell of cell constant

1.15 c m^{- 1}

. The equivalent conductance will be:

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The resistance of a 0.1 N solution of a salt is found to be 2.5×103 ohm. The equivalent conductance of the solution is: (Cell constant = 1.15 cm−1)

The correct option is B 4.6 Scm2/eqConductivity=conductance×cell constant=12.5×103×1.15=4.6×10−4 Scm−1Λeq=κ×1000N=4.6×10−4×10000.1Λeq=4.6 Scm2/eq

The resistance of a 0.1 $N$ solution of a salt is found to be $2.5 \times 10^{3}$ $o h m$ . The equivalent conductance of the solution is: (Cell constant = $1.15 c m^{- 1}$ )

The correct option is B $4.6 S c m^{2} / e q$
$Conductivity = conductance \times cell constant = \frac{1}{2.5 \times 10^{3}} \times 1.15 = 4.6 \times 10^{- 4} S c m^{- 1} Λ_{e q} = \frac{κ \times 1000}{N} = \frac{4.6 \times 10^{- 4} \times 1000}{0.1} Λ_{e q} = 4.6 S c m^{2} / e q$