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What concentration of N2 should be present in a glass of water at room temperature assume temperature of 25 C a total pressure of 1atm and mole fraction of nitrogen in air is 0.78 and Henry constant is 8.42 M/mmHg?
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What concentration of N2 should be present in a glass of water at room...
Concentration of N2 in a Glass of Water at Room Temperature

Given Information:
- Temperature (T) = 25°C = 298 K
- Total pressure (P) = 1 atm
- Mole fraction of nitrogen in air (X) = 0.78
- Henry's constant (H) = 8.42 M/mmHg

Explanation:

The concentration of a gas dissolved in a liquid can be determined using Henry's law. According to Henry's law, the concentration of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid. Mathematically, it can be expressed as:

C = H * P

Where:
C = Concentration of the gas in the liquid
H = Henry's constant
P = Partial pressure of the gas above the liquid

In this case, we need to find the concentration of nitrogen (N2) in a glass of water at room temperature.

Step 1: Convert the given temperature from Celsius to Kelvin:
T(°C) + 273 = T(K)
25 + 273 = 298 K

Step 2: Calculate the partial pressure of nitrogen (N2) in the air using the mole fraction:
Partial pressure of N2 (P_N2) = X * P
P_N2 = 0.78 * 1 atm
P_N2 = 0.78 atm

Step 3: Convert the partial pressure of nitrogen from atm to mmHg:
1 atm = 760 mmHg
Partial pressure of N2 (P_N2) = 0.78 * 760 mmHg
P_N2 = 608 mmHg

Step 4: Calculate the concentration of N2 in water using Henry's law equation:
C_N2 = H * P_N2
C_N2 = 8.42 M/mmHg * 608 mmHg

Step 5: Convert the concentration of N2 from M/mmHg to M:
1 M/mmHg = 1 M
C_N2 = 8.42 M

Conclusion:

The concentration of N2 in a glass of water at room temperature (25°C) with a total pressure of 1 atm and a mole fraction of nitrogen in air of 0.78 is 8.42 M.
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