Consider the strong electrolytes Z m X n ,   U m Y p and V m X n . Limiting molar conductivity…

Consider the strong electrolytes ZmXn, UmYp and VmXn. Limiting molar conductivity \(\left(\wedge^{\circ}\right)\) of UmYp and VmXn are 250 and 440 S cm2 mol-1, respectively. The value of m+n+p is

Given:

Ion Zn+ Up+ Vn+ Xm- Ym-
λ0S cm 2mol-1 50.0 25.0 100.0 80.0 100.0

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.

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

ZmXnmZn++nXm-

  50m+80n=340

5m+8n=34         ...i

UmYpmUp++pYm-

25 m+100p=λm0UmYp=250

m+4p=10     ...ii

VmXnmVn++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

Asked in: JEE Advanced 2022 (Paper 2)

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