A quantity of electricity liberates 3.6g of Silver from its salt. What mass of aluminium Will be liberated from its salt by the same quantity of electricity? [Al = 27, Ag = 108].
2.7g
1.2g
0.9g
0.3g
Explanation
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1F--108g of silver
xF---3.6g of silver
Where F=96500
So,
96500*1----108g
xF------------3.6g
Cross multiply
96500*3.6=108x
Make x the subject of the formula
x=96500*3.6/108
X=3216.67coloumb
So,
3F---27g of aluminium
According to the question,it say the same quantity of electricity, which is the same as that of silver, which is 3216.67coloumbs of electricity.
Therefore,
3F--------------27g of Al
3216.62----xg of Al
Where F=96500
3*96500---27
3216.67---x
By cross multiplying,we have
3*96500x=3216.67*27
Making x the subject of the formula we have
X=3216.67*27/289500
X=0.3g of Al
So the correct answer is D
ncAg = ncAl
n= mole, c= charge
n= m/M
(m/M)cAg = (m/M)Al
(3.6/108)×1 = (m/27)×3
3.6/108= 3m/27
324m=97.2
m= 97.2/324
m= 0.3g of Al
my answer is C
96500C liberates 108g of Ag
xc will liberate 96500*3.6/108
= 3216.6C
therefore using this same quantity of electricity
96500C liberates 27g of Al
3216.67 will liberate 3216.67*27/96500
= 0.9g of Al
m1/E1 = m2/E2
E = atomic mass ÷ valence
E1 = 27/3, E2 = 108/1
3.6/108 = m2/9
m2 = (3.6x9)/108
m2 = 0.3
mass of silver = 3.6g
mass of Al = ?
molar mass of Ag = 108g/mol
molar mass of Al = 27g/mol
using the formula Q= nF
Q= quantity of charge
n= no of moles
f= faraday constant (96500)
let's find no of moles for Ag
n=m/mm
n=3.6/108
n= 0.0333mol
Q= nF
Ag carries a charge of +1
Q= 0.0333* 96500
Q= 3216.67C
the question says by the same quantity of electricity
so lets find the no of moles for Al by using Q= nF
n= Q/F
Alluminium carries a charge of +3
n= 3216.67/3*96500
n= 3216.67/289500
n= 0.0111mol
n=mass/molarmass
mass = n*mm
mass= 0.0111*27
mass= 0.3g...
I don't understand,pls I need a detailed explanation on it and I want to know the formula
you can also use a the method explained in the video but with formula for clarity.
Remember:
m = ZQ
Q = m/Z
where Z =( Atomic Mass) ÷ 96500 x Valency
:• Q of Ag = (valency * 96500 * given mass)/Atomic mass
Q of Ag =( 1 * 96500 * 3.6)/108
Q of Ag = 3217C
use the Charge of Ag for AL.
Q of Ag = (nF*m)÷Atomic mass
3217 = (3*9650m) ÷ 27.
make m subject of formula
m =(9*3217)÷96500
m = 0.3g
The answer is D.
(It)/cf =m/Mm
For silver, Ag+
It/1*96500=3.6/27
It= (96500*3.6)/27
It= 12866.67
For Al.
It/cf=m/mm
12866.67/3*96500=m/27
m=(27*12866.67)/(3*96500)
M=1.2g
lol first trial and i got c simply using my secondary School knowledge
m÷mm=m÷mm
where m1=3.6g
m2=unknown
mm1(silver)=108
mm2(aluminum)=27
3.6/108=m2/27
cross multiply
m2=3.6×27/108=0.9g
which is option C
