Question

The solubility of gas in a liquid is profoundly affected by pressure. This relation is expressed by Henry's law, which states that the solubility of a gas in a liquid, expressed in moles per liter (or M), at a given temperature is directly proportional to the partial pressure of the gas over the solution. This relation can be expressed mathematically as solubility=k⋅P.The constant k is characteristic of the specific gas and P is the partial pressure of the gas over the solution usually expressed in atmospheres. The constant k usually has units of moles per liter per atmosphere and is reported at 25 ∘C. This expression can be used to calculate any of the three variables provided that the other two are known for any gas.1) Air is a mixture of gases that is about 78.0% N2 by volume. When air is at standard pressure and 25.0 ∘C, theN2 component will dissolve in water with a solubility of 4.88×10−4 M. What is the value of Henry's law constant for N2 under these conditions?2) The vapor pressure of benzene, C6H6, is 100.0 torr at 26.1 ∘C. Assuming Raoult’s law is obeyed, how many moles of a nonvolatile solute must be added to 100.0 mL of benzene to decrease its vapor pressure by 10.0% at 26.1 ∘C? The density of benzene is 0.8765 g/cm3.3) A 2.650×10−2M solution of NaCl in water is at 20.0∘C. The sample was created by dissolving a sample of NaClin water and then bringing the volume up to 1.000 L. It was determined that the volume of water needed to do this was 999.3 mL . The density of water at 20.0∘C is 0.9982 g/mL.

Answer 1

Answer:

1) 6.25 × 10⁻⁴ M atm⁻¹

2) 0.125 mol

Explanation:

1) Air is a mixture of gases that is about 78.0% N₂ by volume. When air is at standard pressure and 25.0 °C, the N₂ component will dissolve in water with a solubility of 4.88×10−4 M. What is the value of Henry's law constant for N₂ under these conditions?

When air is at standard pressure (1.00 atm), the partial pressure of N₂ depends on its proportion. If N₂ is 78.0% by volume, its partial pressure will be:

pN₂ = 0.780 × 1.00 atm = 0.780 atm

According to Henry's Law:

S = k × P

k = S / P = 4.88 × 10⁻⁴ M / 0.780 atm = 6.25 × 10⁻⁴ M atm⁻¹

2) The vapor pressure of benzene, C₆H₆, is 100.0 torr at 26.1 °C. Assuming Raoult’s law is obeyed, how many moles of a nonvolatile solute must be added to 100.0 mL of benzene to decrease its vapor pressure by 10.0% at 26.1 ∘C? The density of benzene is 0.8765 g/cm³.

According to Raoult’s law,

Δp = p(solvent) × x(solute)

where,

Δp is the decrease in the vapor pressure of the solvent

p(solvent) is the vapor pressure of the pure solvent

x(solute) is the mole fraction of the solute

Δp is 10.0% of 100.0 torr, that is, 10.0 torr. p(solvent) is 100.0 torr. Then,

$\Delta p=p(solvent) \times x(solute)\\x(solute)=\frac{\Delta p}{p(solvent)} =\frac{10.0torr}{100.0torr} =0.100$

In 100.0 mL of benzene, the number of moles is:

$100.0mL.\frac{0.8765g}{mL} .\frac{1mol}{78.11g} =1.122mol$

Finally,

$x(solute)=0.100=\frac{n_{solute}}{n_{solute}+n_{solvent}} =\frac{n_{solute}}{n_{solute}+1.122mol} \\0.100.n_{solute}+0.1122mol=n_{solute}\\n_{solute}=0.125mol$