Explanation:
The given data is as follows.
Current (I) = 3.50 amp, Mass deposited = 100.0 g
Molar mass of Cr = 52 g
It is known that 1 faraday of electricity will deposit 1 mole of chromium. As 1 faraday means 96500 C and 1 mole of Cr means 52 g.
Therefore, 100 g of Cr will be deposited by "z" grams of electricity.
z =
= 185576.9 C
As we know that, Q = I × t
Hence, putting the given values into the above equation as follows.
Q = I × t
185576.9 C =
t = 53021.9 sec
Thus, we can conclude that 100 g of Cr will be deposited in 53021.9 sec.
MAl₂(SO₄)₃·xH₂O:
(mAl×2) + (mS×3) + (mO×12) + (mH₂O×x)
(27×2)+(32×3)+(16×12)+(x×18) = 342 + 18x [g]
mAl₂: 27×2 = 54 [g]
54g ---------- 13,63%
342+18x ---- 100%
0,1363(342+18x) = 54
46,6146 + 2,4534x = 54
2,4534x = 7,3854
x ≈ 3
>>> Al₂(SO₄)₃·3H₂O <<<<
:)
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