When 3.86 amperes current is passed through an electrolyte for 50 minutes, 2.4 grams of a covalent metal is deposited. The gram atomic mass of the metal ( in grams) is:
1 point
24
12
64
40
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Answer:
The quantity of electricity passed =Q(C)=I(A)×t(s)=3.86A×3000s=11580C.
The gram atomic weight of divalent metal = W
mol
g
Moles of electrons passed =
96500
mole
−
C
Q(C)
=
96500
mole
−
C
11580C
=
96500
11580
mole
−
The mass of divalent metal deposited = 2.4 g =atomicweightofmetal×moleratio×molesofelectronspassed
=W
mol
g
×
2mole
−
1molmetal
×
96500
11580
mole
−
W=40
mol
g
.
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