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Question : 145 of 160
Marks:
+1,
-0
Solution:
Given,
‌I=96.5A‌t=100 s⇒m=? Atomic weight of
Al=27uQ‌=I×t=96.5×100‌=9650CAlCl3→Al3+‌+3Cl−Al3++3e−‌→Al96500×3C of electricity deposits, 1 mole Al.
∴9650C‌ will deposit ‌Al‌=‌=0.0333‌mol.‌ Weight of Al deposited ‌‌=n×‌ molar mass ‌‌=0.0333×27‌≅0.90g
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