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NEET 2026
Exam
NEET
Year
2026
Subject
Chemistry
Chapter
Chapter

NEET 2026 · Chemistry · Chapter

A conductivity cell filled with 0.1 M HCl solution has a resistance...

A conductivity cell filled with 0.1 M0.1\text{ M} HCl\text{HCl} solution has a resistance of 50 Ω50\text{ }\Omega at 300 K300\text{ K}. If the conductivity of 0.1 M0.1\text{ M} HCl\text{HCl} is 3.6×10−2 S cm−13.6 \times 10^{-2}\text{ S cm}^{-1}, what will be the conductivity of a 0.02 M0.02\text{ M} NaCl\text{NaCl} solution if the same cell filled with it offers a resistance of 400 Ω400\text{ }\Omega at the same temperature?

Options

  1. A7.2×10−3 S cm−17.2 \times 10^{-3}\text{ S cm}^{-1}
  2. B4.5×10−3 S cm−14.5 \times 10^{-3}\text{ S cm}^{-1}
  3. C2.5×10−3 S cm−12.5 \times 10^{-3}\text{ S cm}^{-1}
  4. D1.8×10−3 S cm−11.8 \times 10^{-3}\text{ S cm}^{-1}

Correct answer

B. 4.5×10−3 S cm−14.5 \times 10^{-3}\text{ S cm}^{-1}

Step-by-step solution

The cell constant G∗G^{*} is given by the product of conductivity κ\kappa and resistance RR.

For the 0.1 M0.1\text{ M} HCl\text{HCl} solution:

G∗=κ1×R1G^{*} = \kappa_{1} \times R_{1}

G∗=3.6×10−2 S cm−1×50 Ω=1.8 cm−1G^{*} = 3.6 \times 10^{-2}\text{ S cm}^{-1} \times 50\text{ }\Omega = 1.8\text{ cm}^{-1}

Since the same conductivity cell is used, the cell constant remains the same.

For the 0.02 M0.02\text{ M} NaCl\text{NaCl} solution:

κ2=G∗R2\kappa_{2} = \dfrac{G^{*}}{R_{2}}

κ2=1.8 cm−1400 Ω=4.5×10−3 S cm−1\kappa_{2} = \dfrac{1.8\text{ cm}^{-1}}{400\text{ }\Omega} = 4.5 \times 10^{-3}\text{ S cm}^{-1}

Answer: 4.5×10−3 S cm−14.5 \times 10^{-3}\text{ S cm}^{-1}
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