The resistance of $1 N$ solution of $C H_{3} C O O H$ is $250 o h m$ when measured in a cell of cell constant $1.15 c m^{- 1}$ . The equivalent conductance will be:

4.6 o h m^{- 1} c m^{2} e q^{- 1}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

9.2 o h m^{- 1} c m^{2} e q^{- 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

18.4 o h m^{- 1} c m^{2} e q^{- 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.023 o h m^{- 1} c m^{2} e q^{- 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $4.6 o h m^{- 1} c m^{2} e q^{- 1}$
$K = C \times \frac{l}{A} = \frac{1}{250} \times 1.15 = 4.6 \times 10^{- 3} o h m^{- 1} c m^{- 1}$
$\land_{e} = K \times \frac{1000}{N} = 4.6 \times 10^{- 3} \times \frac{1000}{1} = 4.6 o h m^{- 1} c m^{2} e q^{- 1}$ .

Suggest Corrections

Similar questions

Q. The resistance of 1 N solution of acetic acid is 250 ohm, when measured in a cell of cell constant 1.15

c m^{- 1}

. The equivalent conductance (in

o h m^{- 1} c m^{2} e q^{- 1}

) of 1 N acetic acid is:

Q. The resistance of

1 N

solution of acetic acid is

250 o h m

, when measured in a cell having a cell constant of

1.15 {c m}^{- 1}

. The equivalent conductance (in

{o h m}^{- 1} {c m}^{2} {e q u i v}^{- 1}

) of

1 N

acetic acid is:

Equivalent conductance of 1 M propanoic acid is 10 $o h m^{- 1} c m^{2} e q^{- 1}$ and that at infinite dilution 200 $o h m^{- 1} c m^{2} e q^{- 1}$ . The pH of the propanoic acid solution is:

Q. The resistance of

1 N

solution of acetic acid is

250 Ω

, when measured in a cell having a cell constant of

1.15 {c m}^{- 1}

. The equivalent conductance (in

Ω^{- 1} {c m}^{2} {e q u i v}^{- 1}

) of

1 N

acetic acid is:

The resistance of $1 N$ solution of acetic acid is $250 o h m .$ when measured in a cell of cell constant $1.15$ $c m^{- 1}$ . The equivalent conductance of $1 N$ acetic acid (in $o h m^{- 1}$ $c m^{2}$ $e q v^{- 1}$ ) is:

Join BYJU'S Learning Program

The resistance of 1 N solution of CH3COOH is 250 ohm when measured in a cell of cell constant 1.15 cm−1. The equivalent conductance will be:

The correct option is A 4.6 ohm−1cm2eq−1K=C×lA=1250×1.15=4.6×10−3 ohm−1cm−1∧e=K×1000N=4.6×10−3×10001=4.6 ohm−1cm2eq−1.

The resistance of $1 N$ solution of $C H_{3} C O O H$ is $250 o h m$ when measured in a cell of cell constant $1.15 c m^{- 1}$ . The equivalent conductance will be:

The correct option is A $4.6 o h m^{- 1} c m^{2} e q^{- 1}$
$K = C \times \frac{l}{A} = \frac{1}{250} \times 1.15 = 4.6 \times 10^{- 3} o h m^{- 1} c m^{- 1}$
$\land_{e} = K \times \frac{1000}{N} = 4.6 \times 10^{- 3} \times \frac{1000}{1} = 4.6 o h m^{- 1} c m^{2} e q^{- 1}$ .