Home
Class 12
CHEMISTRY
A metal is know to form fluoride MF(2). ...

A metal is know to form fluoride `MF_(2)`. When `10A` of electricity is passed through a molten sat for `330sec, 1.95g` of metal is deposited. Find the atomic weight of M. what will be the quantity electricity required to deposit the same mass of `Cu` form `CuSO_(4)`?

Text Solution

Verified by Experts

Eq. of metal `=(i.t.)/(96500)` (i=10 ampere, t=330sec)
`(1.95)/(E)=(10 xx 330)/(96500)`
`E_("metal")=57.0`
At. Wt. of metal `=57 xx 2 ("metal is bivalent as salt is "MF_(2))`
=114.0
Also, if `W_(cu)=1.95`g, then
Eq. of Cu `=(i.t.)/(96500)`
or `(1.95)/(63.6//2)=(i.t.)/(96500)`
i.e. =Q=5917.45 coulomb
Promotional Banner

Similar Questions

Explore conceptually related problems

When 0.04 F of electricity is passed through a solution of CaSO_(4) , then the weight of Ca^(2+) metal deposited at the cathode is

3F eletricity was passed through molten FeO. Weight of iron metal (Atomic wt=56) deposited at cathode (in g) is

10800 C of electricity passed through the electrolyte deposited 2.977g of metal with atomic mass 106.4 g mol^(-1) . The charge on the metal cation is

Two faraday of electricity is passed through a solution of CuSO_(4) . The mass of copper deposited at the cathode is: (at mass of Cu = 63.5 amu)

10800 C of electricity passed through an electrolyte deposited 2.977 g of metal with atomic mass 106.4 " g mol"^(-1) The charge on the metal cation is :

A current of 4A is passed through a molten solution for 45 min. 2.977 g of metal is deposited. Calculate the charge carried by the metal cation if its atomic mass is 106.4 g/mol.

When electricity is passed through a solution of AlCl_(3) and 13.5g of Al is deposited, the number of Faraday of electricity passed must be ………………….F a.0.5 b.1.0 c.1.5 d.2.0

When a quantity of electricity is passed through CuSO_(4) solution, 0.16 g of copper gets deposited. If the same quantity of electricity is passed through acidulated water, then the volume of H_(2) liberated at STP will be : (given atomic weight of Cu=64)

On passing 3 A of electricity for 50 min, 1.8 g of metal deposits. The equivalent mass of metal is

When 9.65 coulomb of electricity is passed through a solution of silver nitrate (Atomic mass of Ag = 108 g mol^(-1) , the amount of silver deposited is :