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Partial pressure of N(2) gas at 298 K is...

Partial pressure of `N_(2)` gas at 298 K is 0.987 bargt If is bubbled through water at 298 K, how many millimoles of `N_(2)` gas would be dissolved in 1 litre of water ? (Given : `K_(H)` for `N_(2)` at 298 K =76.48 bar).

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To solve the problem of how many millimoles of \( N_2 \) gas would be dissolved in 1 liter of water at 298 K, we can use Henry's Law, which states that the concentration of a gas in a liquid is proportional to the partial pressure of that gas above the liquid. ### Step-by-Step Solution: **Step 1: Write down Henry's Law** According to Henry's Law: \[ P = K_H \cdot x ...
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If N_(2) gas is bubbled through water at 293 K , how many millimoles of N_(2) gas would dissolve in 1 L of water. Assume that N_(2) exerts a partial pressure of 0.987 bar. Given that Henry law constant for N_(2) at 293 K is 76.48 kbar.

If N_(2) gas is bubbled through water at 293 K , how many millimoles of N_(2) gas would dissolve in 1 L of water. Assume that N_(2) exerts a partial pressure of 0.987 bar. Given that Henry law constant for N_(2) at 293 K is 76.48 kbar.

Knowledge Check

  • The Henry's law constant for the solubility of N_(2) gas in water at 298 K is 1.0 xx 10^(5) atm . The mole fraction of N_(2) in air is 0.8 . The number of moles of N_(2) from air dissolved in 10 moles of water at 298 K and 5 atm . Pressure is:

    A
    `4.0xx10^(-4)`
    B
    `4.0xx10^(-5)`
    C
    `5.0xx10^(-4)`
    D
    `4.0xx10^(-6)`

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