When $96500$ coulombs of electricity are passed through barium chloride solution, the amount of barium deposited will be:

0.5 m o l

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

1.0 m o l

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1.5 m o l

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2.0 m o l

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $0.5 m o l$
${B a S O}_{4} ⇌ {B a}^{2 +} + {S O}_{4}^{2 -}$
$∴$ No. of electrons involved $= n = 2$
$2 \times 96500$ Coulomb deposits one mole of barium
$∴ 96500$ Coilomb deposits $0.5$ mole of barium.

Suggest Corrections

Similar questions

Q. When

96500

coulomb of electricity is passed through a copper sulphate solution, the amount of copper deposited will be:

Q. When

9.65

coulomb of electricity is passed through a solution of silver nitrate, the amount of silver deposited is:

When

96.5

coulomb of electricity is passed through a solution of

A g N O

_{3}

.The amount of

A g

deposited as :

When 9.65 coulombs of electricity is passed through a solution of silver nitrate

(atomic weight of taking as 108) the amount of silver deposited is

[EAMCET 1992; KCET 2000]

When 9.65 coulombs of electricity is passed through a solution of silver nitrate (atomic weight of Ag = 107.3, approximately 108) the amount of silver deposited is:

Join BYJU'S Learning Program

When 96500 coulombs of electricity are passed through barium chloride solution, the amount of barium deposited will be:

The correct option is A 0.5 molBaSO4⇌Ba2++SO2−4∴ No. of electrons involved =n=22×96500 Coulomb deposits one mole of barium∴96500 Coilomb deposits 0.5 mole of barium.

When $96500$ coulombs of electricity are passed through barium chloride solution, the amount of barium deposited will be:

The correct option is A $0.5 m o l$
${B a S O}_{4} ⇌ {B a}^{2 +} + {S O}_{4}^{2 -}$
$∴$ No. of electrons involved $= n = 2$
$2 \times 96500$ Coulomb deposits one mole of barium
$∴ 96500$ Coilomb deposits $0.5$ mole of barium.