An equeous solution of `Na_(2)SO_(4)` was electrolysed for 10 min. 82 ml of a gas was produced at anode and collected over water at `27^(@)C` at a total pressure of 580 torr. Determine the current that was used in amp.
Given: Vapour pressure of `H_(2)O` at `27^(@)C = 10` torr
`R = 0.082 atm L//mol-K`
Write your answer exclusing decimal places.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the partial pressure of the gas produced at the anode
The total pressure of the gas collected is given as 580 torr, and the vapor pressure of water at 27°C is 10 torr. Therefore, the partial pressure of the gas (P) can be calculated as:

\[ P = \text{Total Pressure} - \text{Vapor Pressure} \]
\[ P = 580 \, \text{torr} - 10 \, \text{torr} = 570 \, \text{torr} \]

### Step 2: Convert the pressure from torr to atm
To use the ideal gas law, we need to convert the pressure from torr to atm. The conversion factor is \( 1 \, \text{atm} = 760 \, \text{torr} \).

\[ P = \frac{570 \, \text{torr}}{760 \, \text{torr/atm}} \approx 0.750 \, \text{atm} \]

### Step 3: Convert the volume from mL to L
The volume of the gas produced is given as 82 mL. We need to convert this to liters:

\[ V = \frac{82 \, \text{mL}}{1000} = 0.082 \, \text{L} \]

### Step 4: Use the ideal gas law to find the number of moles of gas produced
The ideal gas law is given by the equation \( PV = nRT \), where:
- \( P \) is the pressure in atm,
- \( V \) is the volume in liters,
- \( n \) is the number of moles,
- \( R \) is the universal gas constant (0.082 atm L/mol K),
- \( T \) is the temperature in Kelvin.

First, convert the temperature from Celsius to Kelvin:

\[ T = 27 + 273 = 300 \, \text{K} \]

Now, substituting the values into the ideal gas law:

\[ n = \frac{PV}{RT} \]
\[ n = \frac{(0.750 \, \text{atm})(0.082 \, \text{L})}{(0.082 \, \text{atm L/mol K})(300 \, \text{K})} \]

Calculating \( n \):

\[ n = \frac{0.0615}{24.6} \approx 0.0025 \, \text{mol} \]

### Step 5: Calculate the charge using Faraday's law
According to Faraday's law, the charge (Q) is given by:

\[ Q = n \times F \]

Where \( F \) (Faraday's constant) is approximately \( 96500 \, \text{C/mol} \). Since the reaction at the anode produces oxygen gas (O2), which involves the transfer of 4 electrons:

\[ Q = n \times F \]
\[ Q = 0.0025 \, \text{mol} \times 96500 \, \text{C/mol} \]
\[ Q \approx 241.25 \, \text{C} \]

### Step 6: Calculate the current (I)
The current can be calculated using the formula:

\[ I = \frac{Q}{t} \]

Where \( t \) is the time in seconds. The time given is 10 minutes, which we convert to seconds:

\[ t = 10 \, \text{min} \times 60 \, \text{s/min} = 600 \, \text{s} \]

Now substituting the values:

\[ I = \frac{241.25 \, \text{C}}{600 \, \text{s}} \approx 0.402 \, \text{A} \]

### Step 7: Write the final answer excluding decimal places
The question asks for the answer excluding decimal places. Therefore, we take the integer part of the current:

**Final Answer: 0**