Q. No.Cl- = 71 S cm2mol–1, then the limiting molar conductivity of CaCl2 is:
1. 341 S cm2mol–1
2. 261 S cm2mol–1
3. 126 S cm2mol–1
4. 431 S cm2mol–1
| Column I | Column II | ||
| (A) | \( \wedge_m\) | (i) | \(l/A\) |
| (B) | \(G^*\) | (ii) | \(\rho l/A \) |
| (C) | \(\kappa\) | (iii) | \(\kappa /C\) |
| (D) | R | (iv) | \(\frac{G^*}{R}\) |
1. (A) →(i), (B)→(iii), (C) →(ii), (D)→ (iv)
2. (A) →(iii), (B)→(i), (C) →(iv), (D)→ (ii)
3. (A) →(ii), (B)→(iv), (C) →(iii), (D)→ (i)
4. (A) →(iv), (B)→(ii), (C) →(i), (D)→ (iii)
| 1. | \(124.66 ~ \text {Scm}^2/ \text {mol} \) | 2. | \(135.66 ~ \text {Scm}^2/ \text {mol} \) |
| 3. | \(198.20 ~ \text {Scm}^2/ \text {mol} \) | 4. | \(175.78 ~ \text {Scm}^2/ \text {mol} \) |
Calculate the equivalent conductance of C₂H₅COOH at infinite dilution if the equivalent conductivities of NaCl, HCl and C₂H₅COONa at infinite dilution are 126.45, 426.16 and 91 Ω⁻¹ cm² respectively.
1. 201.28
2. 390.71
3. 698.28
4. 540.48
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1. 1.29 cm–1
2. 12.9 cm–1
3. 129 cm–1
4. 1290 cm–1
Match the terms given in Column I with the units given in Column II.
|
Column I |
Column II |
||
| A. | 1. |
S cm–1 |
|
| B. | Ecell | 2. | m–1 |
| C. | κ | 3. | S cm2 mol–1 |
| D. | G* | 4. | V |
Codes:
| Options: | A | B | C | D |
| 1. | 3 | 4 | 1 | 2 |
| 2. | 1 | 2 | 3 | 5 |
| 3. | 5 | 4 | 3 | 2 |
| 4. | 4 | 5 | 3 | 2 |
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1. \(124~ \Omega ^{-1} ~cm^2~mol^{-1}\)
2. \(204~ \Omega ^{-1} ~cm^2~mol^{-1}\)
3. \(101~ \Omega ^{-1} ~cm^2~mol^{-1}\)
4. \(300~ \Omega ^{-1} ~cm^2~mol^{-1}\)
The correct choice among the given regarding formula of Conductivity is -
(a)
(b)
(c)
(d)
1. (a, b)
2. (b, c)
3. (c, d)
4. (a, d)
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The molar conductivity of a solution of AgNO3 (considering the given data for this solution) at 298 K is:
(a) Concentration: 0.5 mol/dm3
(b) Electrolytic conductivity: 5.76×10–3 S cm–1
| 1. | 2.88 S cm2/mol | 2. | 11.52 S cm2/mol |
| 3. | 0.086 S cm2/mol | 4. | 28.8 S cm2/mol |
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1. 203
2. 421
3. 143
4. 303
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