$10800 C$ of electricity on passing through the electrolyte solution deposited $2.977$ of metal with atomic mass ${106.4 g m o l}^{- 1}$ the charge on the metal cation is:

+ 4

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

+ 3

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

+ 2

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

+ 1

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

Open in App

Solution

The correct option is A $+ 4$
$10800 C \to 2.977 g$ of metal deposited.
$96500 C \to \frac{2.977}{10800} \times 96500 = 26.600 g$
Charge on metal cation is $= \frac{106.4}{26.6}$
$= + 4$

Suggest Corrections

Similar questions

10800 C

of electricity through the electrolyte deposited

2.977 g

of metal with atomic mass

106.4 g m o l^{- 1}

. The charge on the metal cation is

10800 C

of electricity through the electrolyte deposited

2.977 g

of metal with atmic mass

106.4 g m o l^{- 1}

. The charge on the metal cation is:

10800C ofelectricity through the electrolyte deposited 2. 977 gram of metal with atomic mass of 106.4 gram per mole find the charge on the metal cations

Q. A current of

3 A

was passed for 1 hour through an electrolyte solution of

A_{x} B_{y}

in water. If

2.977

g of A (atomic wright

106.4

) was deposited at the cathode and

B

was a monovalent ion, the formula of electrolyte was:

Q. 3.80 gm of metal M is deposited at the cathode by passing 0.2 F of electricity through the solution of metal chlorides. The formula of the metal chloride is:
(atomic mass of M = 57 g/mol)

Join BYJU'S Learning Program

10800C of electricity on passing through the electrolyte solution deposited 2.977 of metal with atomic mass 106.4gmol−1 the charge on the metal cation is:

The correct option is A +410800C→2.977g of metal deposited.96500C→2.97710800×96500=26.600gCharge on metal cation is =106.426.6 =+4

$10800 C$ of electricity on passing through the electrolyte solution deposited $2.977$ of metal with atomic mass ${106.4 g m o l}^{- 1}$ the charge on the metal cation is:

The correct option is A $+ 4$
$10800 C \to 2.977 g$ of metal deposited.
$96500 C \to \frac{2.977}{10800} \times 96500 = 26.600 g$
Charge on metal cation is $= \frac{106.4}{26.6}$
$= + 4$