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JEE Advanced 2022Paper 2
Exam
JEE Advanced
Year
2022
Paper
Paper 2
Subject
Chemistry
Chapter
Electrochemistry

JEE Advanced 2022 · Paper 2 · Chemistry · Electrochemistry

Consider the strong electrolytes Z _ m X _ n , U _ m Y _ p and V _...

Consider the strong electrolytes ZmXn,UmYpandVmXn{Z}_{m}{X}_{n}, {U}_{m}{Y}_{p} and {V}_{m}{X}_{n}. Limiting molar conductivity of UmYpandVmXn{U}_{m}{Y}_{p} and {V}_{m}{X}_{n} are 250and440Scm2mol−1250 and 440S{cm}^{2}{mol}^{- 1}, respectively. The value of \left(m + n + p\right)is</p><p>Given:</p><tableborder="1"cellpadding="1"cellspacing="1"style="width:500px;"><tbody><tr><td>Ion</td><td>is</p><p>Given:</p><table border="1" cellpadding="1" cellspacing="1" style="width: 500px;"> <tbody> <tr> <td>Ion</td> <td>{Z}^{n +}</td><td></td> <td>{U}^{p +}</td><td></td> <td>{V}^{n +}</td><td></td> <td>{X}^{m -}</td><td></td> <td>{Y}^{m -}</td></tr><tr><td></td> </tr> <tr> <td>{\lambda}^{0}\left(Scm 2 {mol}^{- 1}\right) 50.050 . 0 25.025 . 0 100.0100 . 0 80.080 . 0 100.0100 . 0

λ0isthe{\lambda}^{0}is thelimiting molar conductivity of ions

The plot of molar conductivity \left(Λ\right)ofZmXnof {Z}_{m}{X}_{n} vs cc\dfrac{1}{2} is given below.

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

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

Correct answer

0

Step-by-step solution

λm=λm°−AC{\lambda}_{m} = {\lambda}_{m}^{°} - A\sqrt{C}

For electrolyte ZmXn{Z}_{m}{X}_{n} and from given curve

λm(ZmXn)=λm0(ZmXn)−AC{\lambda}_{m}\left({Z}_{m}{X}_{n}\right) = {\lambda}_{m}^{0}\left({Z}_{m}{X}_{n}\right) - A\sqrt{C}

−A=336−3390.04−0.01=−30.03- A = \dfrac{336 - 339}{0 . 04 - 0 . 01} = - \dfrac{3}{0 . 03}

⇒A=100⇒ A = 100

∴∴ For λm=336Scm2mol−1{\lambda}_{m} = 336S{cm}^{2}{mol}^{- 1}

⇒336=λm0(ZmXn)−100×0.04⇒ 336 = {\lambda}_{m}^{0}\left({Z}_{m}{X}_{n}\right) - 100 \times 0 . 04

λm0=336+4=340Scm2mol−1{\lambda}_{m}^{0} = 336 + 4 = 340S{cm}^{2}{mol}^{- 1}

ZmXn⟶mZn++nXm−{Z}_{m}{X}_{n} ⟶ {mZ}^{n +} + {nX}^{m -}

∴50m+80n=340∴ 50m + 80n = 340

⇒5m+8n=34...(i)⇒ 5m + 8n = 34 . . . \left(i\right)

UmYp⟶mUp++pYm−{U}_{m}{Y}_{p} ⟶ {mU}^{p +} + {pY}^{m -}

∴25m+100p=λm0(UmYp)=250∴ 25m + 100p = {\lambda}_{m}^{0}\left({U}_{m}{Y}_{p}\right) = 250

⇒m+4p=10...⇒ m + 4p = 10 . . .\left(ii\right)</p><p></p><p>{V}_{m}{X}_{n} ⟶ {mV}^{n +} + {nX}^{m -}</p><p></p><p>∴ 100m + 80n = 440</p><p></p><p>⇒ 5m + 4n = 22 . . . \left(iii\right)

From equation \left(i\right)andand\left(iii\right)

n=3n = 3

m=2m = 2

From equation (ii)</p><p>\left(ii\right)</p><p>p = 2</p><p></p><p>∴ m + n + p = 2 + 3 + 2 = 7</p></blockquote></p></blockquote>

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