The molar conductivity of a $0.5 m o l / d m^{3}$ solution of $A g N O_{3}$ with electrolytic conductivity of $5.76 \times 10^{- 3} S c m^{- 1}$ at $298 K$ is:

2.88 S c m^{2} / m o l

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

11.52 S c m^{2} / m o l

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

0.086 S c m^{2} / m o l

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

28.8 S c m^{2} / m o l

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

Open in App

Solution

The correct option is D $11.52 S c m^{2} / m o l$
Molar Conductivity $= λ_{m} = \frac{k \times 1000}{M o l a r i t y}$
$= \frac{5.76 \times 10^{- 3} S c m^{- 1} \times 1000}{0.5 m o l c m^{- 3}}$
$= 11.52 S c m^{2} / m o l$

Suggest Corrections

Similar questions

Q. The molar conductivity of a

0.5 m o l / d m^{3}

solution of

A g N O_{3}

with electrolytic conductivity of

5.76 \times 10^{- 3} S c m^{- 1}

at 298 K is

The molar conductivity of a $0.5 m o l / d m^{3}$ solution of $A g N O_{3}$ with electrolytic conductiviy of $5.76 \times 10^{- 3} S c m^{- 1}$ at 298 K is

Q. The molar conductivity of a 0.5 mol/dm

^{3}

solution of

A g N O_{3}

with electrolytic conductivity of

5.76 \times 10^{- 3} S c m^{- 1}

at 298 K is :

The molar conductivity of a $0.5 m o l / d m^{3}$ solution of $A g N O_{3}$ with electrolytic conductiviy of $5.76 \times 10^{- 3} S c m^{- 1}$ at 298 K is

Q. Electrolytic conductivity of 0.02 mol/l solution of kcl at 298k is 2.48x10^-2 ohm^-1 cm^-1. Calculate it's molar conductivity.

Join BYJU'S Learning Program

The molar conductivity of a 0.5 mol/dm3 solution of AgNO3 with electrolytic conductivity of 5.76×10−3 Scm−1 at 298 K is:

The correct option is D 11.52 S cm2/molMolar Conductivity=λm=k×1000Molarity=5.76×10−3Scm−1×10000.5molcm−3=11.52Scm2/mol

The molar conductivity of a $0.5 m o l / d m^{3}$ solution of $A g N O_{3}$ with electrolytic conductivity of $5.76 \times 10^{- 3} S c m^{- 1}$ at $298 K$ is:

The correct option is D $11.52 S c m^{2} / m o l$
Molar Conductivity $= λ_{m} = \frac{k \times 1000}{M o l a r i t y}$
$= \frac{5.76 \times 10^{- 3} S c m^{- 1} \times 1000}{0.5 m o l c m^{- 3}}$
$= 11.52 S c m^{2} / m o l$