Question

question 4
indicate the most important type of intermolecular attraction responsible for solvation in each of the following solutions.

solution	intermolecular attraction
toluene (c₇h₈) in benzene (c₆h₆)	--
cacl₂ in water	--

fill in the blanks one option from
part 1: 3 points
part 2: 3 points
part 3: 3 points
9 points
question 5
the henry’s law constant for o₂ is 1.3 × 10⁻³ m/atm at 25 °c. what mass of oxygen would be dissolved in a 40-l aquarium at 25 °c, assuming an atmospheric pressure of 1.00 atm, and that the partial pressure of o₂ is 0.21 atm?
g o₂
type numbers in
10 points

Explanation:

Step1: Calculate the concentration of \(O_2\) using Henry's law

Henry's law is \(C = kP\), where \(C\) is the concentration, \(k\) is Henry's law constant, and \(P\) is the partial pressure.
Given \(k = 1.3\times10^{- 3}\text{ M/atm}\) and \(P = 0.21\text{ atm}\)
\(C=(1.3\times 10^{-3}\text{ M/atm})\times(0.21\text{ atm}) = 2.73\times10^{-4}\text{ M}\)

Step2: Calculate the number of moles of \(O_2\)

Molarity \(M=\frac{n}{V}\), where \(n\) is the number of moles and \(V\) is the volume.
Given \(V = 40\text{ L}\) and \(M=2.73\times 10^{-4}\text{ M}\)
\(n = M\times V=(2.73\times10^{-4}\text{ mol/L})\times(40\text{ L})=0.01092\text{ mol}\)

Step3: Calculate the mass of \(O_2\)

The molar mass of \(O_2\) is \(M_{O_2}=32\text{ g/mol}\)
Using the formula \(m = n\times M\)
\(m=(0.01092\text{ mol})\times(32\text{ g/mol}) = 0.34944\text{ g}\approx0.35\text{ g}\)

Answer:

\(0.35\)