`821mLN_(2) (g)` was collected over liquid water at 300 K and 1 atm.If vapour pressure of `H_(2)O` is 30 torr then moles of `N_(2) (g)` in moist gas mixture is :

Options

A. 0.39
B. 0.032
C. 0.96
D. 0.0013

Explanation

To find the moles of nitrogen gas (N₂) in the moist gas mixture, we can follow these steps:

### Step 1: Calculate the pressure of nitrogen gas
The total pressure (P_total) when the gas is collected over water is equal to the sum of the partial pressures of the gas and the vapor pressure of water.

Given:
- Total pressure (P_total) = 1 atm = 760 torr
- Vapor pressure of water (P_H₂O) = 30 torr

Using the formula:
\[ P_{N_2} = P_{total} - P_{H_2O} \]

Substituting the values:
\[ P_{N_2} = 760 \, \text{torr} - 30 \, \text{torr} = 730 \, \text{torr} \]

### Step 2: Convert the pressure of nitrogen gas to atm
To use the ideal gas law, we need the pressure in atm:
\[ P_{N_2} = \frac{730 \, \text{torr}}{760 \, \text{torr/atm}} \approx 0.9605 \, \text{atm} \]

### Step 3: Convert the volume of nitrogen gas to liters
The volume of nitrogen gas is given as 821 mL. We convert this to liters:
\[ V = \frac{821 \, \text{mL}}{1000} = 0.821 \, \text{L} \]

### Step 4: Use the ideal gas equation to find the moles of nitrogen gas
The ideal gas equation is:
\[ PV = nRT \]

Where:
- P = pressure in atm
- V = volume in liters
- n = number of moles
- R = ideal gas constant = 0.0821 L·atm/(K·mol)
- T = temperature in Kelvin

Rearranging the equation to solve for n:
\[ n = \frac{PV}{RT} \]

Substituting the known values:
\[ n = \frac{(0.9605 \, \text{atm})(0.821 \, \text{L})}{(0.0821 \, \text{L·atm/(K·mol)})(300 \, \text{K})} \]

### Step 5: Calculate n
Calculating the numerator:
\[ 0.9605 \times 0.821 \approx 0.787 \]

Calculating the denominator:
\[ 0.0821 \times 300 \approx 24.63 \]

Now, substituting these values back into the equation:
\[ n \approx \frac{0.787}{24.63} \approx 0.032 \, \text{moles} \]

### Final Answer
The moles of nitrogen gas (N₂) in the moist gas mixture is approximately **0.032 moles**.

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