Chemistry - Electrochemistry Question with Solution | TestHub

ChemistryElectrochemistryConductance/conductivity/^mMedium2 minPYQ_2022
ChemistryMediumnumerical

Consider the strong electrolytes ZmXn, UmYp and VmXn. Limiting molar conductivity of UmYp and VmXn are 250 and 440 S cm2 mol-1, respectively. The value of m+n+p is

Given:

IonZn+Up+Vn+Xm-Ym-
λ0S cm 2mol-150.025.0100.080.0100.0

λ0 is the limiting molar conductivity of ions

The plot of molar conductivity Λ of ZmXn vs c12 is given below.

If the numerical value has more than two decimal places, truncate/round-off the value to TWO decimal places.

Answer:
7.00
Solution:

λm=λm°-AC

For electrolyte ZmXn and from given curve

λmZmXn=λm0ZmXn-AC

-A=336-3390.04-0.01=-30.03

⇒A=100

∴ For λm=336 S cm2 mol-1

⇒336=λm0ZmXn-100×0.04

λm0=336+4=340 S cm2 mol-1

ZmXn⟶mZn++nXm-

∴  50m+80n=340

⇒5m+8n=34         ...i

UmYp⟶mUp++pYm-

∴25 m+100p=λm0UmYp=250

⇒m+4p=10     ...ii

VmXn⟶mVn++nXm-

∴100m+80n=440

⇒5m+4n=22      ...iii

From equation i and iii

n=3

m=2

From equation ii

p=2

∴  m+n+p=2+3+2=7

Stream:JEE_ADVSubject:ChemistryTopic:ElectrochemistrySubtopic:Conductance/conductivity/^m
⏱ 2mℹ️ Source: PYQ_2022

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