Chemistry, 17.01.2020 02:31 tommyaberman

The henry's law constant (kh) for o2 in water at 20°c is 1.28 × 10−3 mol/(l·atm). (a) how many grams of o2 will dissolve in 4.00 l of h2o that is in contact with pure o2 at 1.00 atm? g o2 (b) how many grams of o2 will dissolve in 4.00 l of h2o that is in contact with air where the partial pressure of o2 is 0.209 atm?

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Answer from: destiny2521

0.164 grams of oxygen gas was dissolved in 4.00 Liters of water.

0.0343 grams of oxygen gas was dissolved in 4.00 Liters of water.

Explanation:

Henry's law states that the solubility of a gas in a given volume of a liquid is directly proportional to the partial pressure of that gas above the liquid.

$S=k_H\times p_o$

Where:

S = Solubility of gas in liquid

k_H = Henry's law constant

p_o = Partial pressure of a gas

Given:

The Henry's law constant for oxygen gas in water at 20°C is $1.28 \times 10^{-3} mol/(Latm)$

To find:

a) Mass of oxygen gas in 4.00 Liter of water with pure oxygen at 1.00 atm.

b) Mass of oxygen gas in 4.00 Liter of water with oxygen gas at a partial pressure of 0.209 atm.

Solution:

The Henry's law constant for oxygen gas in water at 20°C = k_H = $1.28 \times 10^{-3} mol/(Latm)$

The pressure of pure oxygen gas above water = p_o=1.00 atm

The solubility of the oxygen gas in water:

$S=1.28\times 10^{-3} mol/(Latm)\times 1.00 atm\\S=1.28\times 10^{-3} mol/L$

There are $1.28\times 10^{-3}$ moles of oxygen gas in 1 liter of water

The volume of the water = 4.00 L

Moles of oxygen gas in 4.00 L of water:

$=1.28\times 10^{-3}\times 4.00 mol=5.12\times 10^{-3}mol$

Mass of $5.12\times 10^{-3}$ moles of oxygen gas:

$=5.12\times 10^{-3}mol\times 31.998 g/mol=0.164 g$

0.164 grams of oxygen gas was dissolved in 4.00 Liters of water.

The Henry's law constant for oxygen gas in water at 20°C = k_H = $1.28 \times 10^{-3} mol/(Latm)$

The partial pressure of oxygen gas above water = p_o=0.209atm

The solubility of the oxygen gas in water:

$S=1.28\times 10^{-3} mol/(Latm)\times 0.209atm\\S=2.68\times 10^{-4} mol/L$

There are $2.68\times 10^{-4}$ moles of oxygen gas in 1 liter of water

The volume of the water = 4.00 L

Moles of oxygen gas in 4.00 L of water:

$=2.68\times 10^{-4}\times 4.00 mol=1.072\times 10^{-3}mol$

Mass of $1.072\times 10^{-3}$ moles of oxygen gas:

$=1.072\times 10^{-3}mol\times 31.998 g/mol=0.0343g$

0.0343 grams of oxygen gas was dissolved in 4.00 Liters of water.

Learn more about Henry's law here:

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Answer from: savannah19dw

Solubility of O₂(g) in 4L water = 3.42 x 10⁻² grams O₂(g)

Explanation:

Graham's Law => Solubility(S) ∝ Applied Pressure(P) => S =k·P

Given P = 0.209Atm & k = 1.28 x 10⁻³mol/L·Atm

=> S = k·P = (1.28 x 10⁻³ mole/L·Atm)0.209Atm = 2.68 x 10⁻³ mol O₂/L water.

∴Solubility of O₂(g) in 4L water at 0.209Atm = (2.68 x 10⁻³mole O₂(g)/L)(4L)(32 g O₂(g)/mol O₂(g)) = 3.45 x 10⁻² grams O₂(g) in 4L water.

Answer from: Quest

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Answer from: Quest

c is the answer because when something is left after something evaporates that shows the chemical change.

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The henry's law constant (kh) for o2 in water at 20°c is 1.28 × 10−3 mol/(l·atm). (a) how many grams...

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