Exams › NEET › Chemistry

If 0.01 M solution of an electrolyte has a resistance of 40 ohms in a cell having a cell constant of 0.4 cm⁻¹, then its molar conductance in ohm⁻¹ cm² mol⁻¹ is:

1. 102
2. 104
3. 10
4. 103

Correct answer: 104

Solution

The molar conductance (Λm) is calculated using the formula Λm = (κ × 1000) / C, where κ = cell constant / resistance. Here, κ = 0.4 / 40 = 0.01 ohm⁻¹ cm⁻¹. Substituting, Λm = (0.01 × 1000) / 0.01 = 104 ohm⁻¹ cm² mol⁻¹.

Related NEET Chemistry questions

• Assertion: When an electric current is passed through the copper sulphate electrolyte, the impure copper anode gradually dissolve and pure copper is deposited on the cathode. Reason: Pure metals are deposited in electroplating techniques.
• What is the electric current required to deposit 0.972 g of chromium in three hours? (E.C.E. of chromium is 0.00018 g/c)
• Which of the following statement (s) differentiate between electrochemical cell and electrolytic cell? This question has multiple correct options
• Among the following, the couple with most positive reduction potential is :
• The resistance of \( 1 \mathrm{N} \) solution of acetic acid is \( 250 \Omega \) when measured in a cell having a cell constant of \( 1.15 \mathrm{cm}^{-1} . \) The equivalent conduction (in oh \( m^{-1} c m^{2} e q u i v^{-1} \) ) of \( 1 N \) acetic acid is
• If \( 9 g m H_{2} O \) is electrolysed completely with the current of \( 50 \% \) efficiency then :

⚔️ Practice NEET Chemistry free + battle 1v1 →