NEET

How much chlorine will be liberated on passing one ampere current for 30 min. through NaCl solution?

0.66 mole
0.33 mole
0.66 g
0.33 g

Solution

The resistance of 0.01 N solution of an electrolyte was found to be 220 ohm at 298 K using a conductivity cell with a cell constant of 0.88 cm^-1. The value of equivalent conductance of solution is –

400 mho cm² g eq^-1
295 mho cm² g eq^-1
419 mho cm² g eq^-1
425 mho cm² g eq^-1

Solution

The unit of equivalent conductivity is

ohm cm
ohm^-1 cm²(g equivalent)^-1
ohm cm²(g equivalent)
S cm^-2

Solution

ohm^-2 cm²(geq)^-1

217.5
390.7
552.7
517.2

Solution

When an electric current of 0.5 ampere is passed through acidulated water for two hours, then volume of hydrogen(H2) at NTP produced will be (1 Coulomb of electricity deposits 0.00001 gm of hydrogen)

0.1 litre
0.6 litre
0.4 litre
0.8 litre

Solution

The highest electrical conductivity of the following aqueous solutions is of

0.1 M difluoroacetic acid
0.1 M fluoroacetic acid
0.1 M chloroacetic acid
0.1 M acetic acid

Solution

Difluoroacetic acid being strongest acid will furnish maximum number of ions showing highest electrical conductivity.

Specific conductance of a 0.1 N KCl solution at 23ºC is 0.012 ohm–1 cm–1. Resistance of cell containing the solution at same temperature was found to be 55 ohm. The cell constant is

0.0616 cm^-1
0.66 cm^-1
6.60 cm^-1
660 cm^-1

Solution

Specific conductance of the solution (k) = 0.012 ohm^-1 cm^-1 and resistance (R) = 55 ohm.
Cell constant = Specific conductance × Observed resistance = 0.012 × 55 = 0.66cm^-1

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