Chemistry-(general)
NEET Chemistry (general) MCQ Question
Type: MCQ-numerical-Medium-Class 12
A steady current of 1.5 amperes was passed through a series of cells containing AgNO3 and CuSO4 until 1.45 g of silver was deposited. How long did the current flow to deposit 1.45 g of silver?
A
10 minutes
B
15 minutes
C
20 minutes
D
25 minutes
Correct Answer
Option B
Detailed Explanation
The mass of silver deposited can be used to calculate the time using the formula: mass = (atomic mass/equivalent mass) * (current * time). Given the atomic mass of Ag is 107.87 g/mol, 1.45 g corresponds to 0.0134 mol, requiring a charge of 96500 C/mol. With a current of 1.5 A, the time is 15 minutes.
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