The number of moles of `Zn^(2+)` ions deposited when a current of `1.5A` is passed for 4 hours through a molten solution of a zinc salt. `(` Assume current efficiency to be `90%)`

Options

A. `6.35`
B. `0.1`
C. `0.4`
D. None of these

Explanation

To solve the problem of determining the number of moles of `Zn^(2+)` ions deposited when a current of `1.5 A` is passed for `4 hours` through a molten solution of a zinc salt with a current efficiency of `90%`, we can follow these steps:

### Step 1: Convert Time to Seconds
First, we need to convert the time from hours to seconds since the current is given in amperes (A), which is coulombs per second (C/s).

\[ \text{Time in seconds} = 4 \text{ hours} \times 3600 \text{ seconds/hour} = 14400 \text{ seconds} \]

### Step 2: Calculate Total Charge (Q)
Using the formula \( Q = I \times t \), where \( I \) is the current in amperes and \( t \) is the time in seconds, we can find the total charge passed through the solution.

\[ Q = 1.5 \text{ A} \times 14400 \text{ s} = 21600 \text{ C} \]

### Step 3: Adjust for Current Efficiency
Since the current efficiency is given as `90%`, we need to adjust the total charge accordingly.

\[ Q_{\text{effective}} = Q \times \text{Efficiency} = 21600 \text{ C} \times 0.90 = 19440 \text{ C} \]

### Step 4: Determine the Charge Required to Deposit Zinc
From electrochemistry, we know that the deposition of `1 mole` of zinc requires `2 moles` of electrons. The charge required for `1 mole` of electrons is `96500 C` (Faraday's constant). Therefore, the charge required for `1 mole` of zinc is:

\[ \text{Charge for 1 mole of Zn} = 2 \times 96500 \text{ C} = 193000 \text{ C} \]

### Step 5: Calculate the Number of Moles of Zinc Deposited
Now we can use the effective charge to find the number of moles of zinc deposited using the unitary method.

\[ \text{Moles of Zn} = \frac{Q_{\text{effective}}}{\text{Charge for 1 mole of Zn}} = \frac{19440 \text{ C}}{193000 \text{ C/mole}} \approx 0.1007 \text{ moles} \]

### Step 6: Round to Significant Figures
Rounding to two significant figures, we find that approximately `0.1 moles` of zinc are deposited.

### Final Answer
The number of moles of `Zn^(2+)` ions deposited is approximately **0.1 moles**.

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