The molar conductivity of a 0-5 mol-dm-3 solution of AgNO-3 with electrolytic conductivity of 5-76 times 10-3- S cm-1- 298 K is -0-086 S cm-2 - mol028-8 S cm-2 - mol2-88 S cm-2 - mol11-52 S cm-2 - mol

Question

Toppr Admin · Accepted Answer

C is -xA0-0-5-xA0-mol-dm3K-5-76-xD7-10-x2212-3Scm-x2212-1T-298K-x3BB-K-xD7-1000m

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 :
$0.086 S c m^{2} / m o l$
$028.8 S c m^{2} / m o l$
$2.88 S c m^{2} / m o l$
$11.52 S c m^{2} / m o l$

C is $0.5 m o l / d m^{3}$
$K = 5.76 \times 10^{- 3} S c m^{- 1}$
T=298K
$λ = \frac{K \times 1000}{m}$

The molar conductivity of a 0.5 mol/dm3 solution of AgNO3 with electrolytic conductivity of 5.76×10−3Scm−1 at 298 K is :0.086 Scm2/mol028.8 Scm2/mol2.88 Scm2/mol11.52 Scm2/mol

C is 0.5 mol/dm3K=5.76×10−3Scm−1T=298Kλ=K×1000m

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 :
$0.086 S c m^{2} / m o l$
$028.8 S c m^{2} / m o l$
$2.88 S c m^{2} / m o l$
$11.52 S c m^{2} / m o l$

C is $0.5 m o l / d m^{3}$
$K = 5.76 \times 10^{- 3} S c m^{- 1}$
T=298K
$λ = \frac{K \times 1000}{m}$