The question is missing the molecules in which the integration ratio of 2:3 will be observed. The complete question is given in the attachment.
Answer:
Molecule (a), (c), and (f) will show two peaks with the integration ratio of 2:3 in their 1H NMR spectrum
Explanation:
In the 1H NMR spectrum, the peak area is dependent on the number of hydrogen in a specific chemical environment. Hence, the ratio of the integration of these signals provides us with the relative number of hydrogen in two peaks. This rationale is used for the assignment of molecules that will give 2:3 integration ratio in the given problem.
- Molecule (a) have two CH₂ and three CH₃ groups. Hence, it will give two peaks and their integration ratio becomes 2:3 (Answer)
- Molecule (b) contains three chemical environments for its hydrogen atoms
- Molecule (c) have a single CH₂ and CH₃ group giving integration ratio of 2:3 (Answer)
- Molecule (d) will give two peaks but their ratio will be 1:3 because of two hydrogens of CH₂ and six hydrogens from two CH₃ groups
- Molecule (e) have three CH and three CH₃ groups, so their ratio will become 1:3
- Molecule (f) contains four CH and two CH₃ groups, giving two peaks. So, the integration ratio of their peaks is 2:3 (Answer)
- Molecules
- (g)
- and
- (h)
- both have two CH and two CH₃ groups giving two peaks with the integration ratio of 1:3
Avagadros number states that 1 mol of any substance is made of 6.022 x 10²³ units.
These units could be atoms that make up an element or molecules that make up a compound. CO₂ is a compound therefore its made of CO₂ molecules.
1 mol of CO₂ contains 6.022 x 10²³ molecules of CO₂
Therefore if 6.022 x 10²³ molecules of CO₂ make up 1 mol
then 1.25 x 10¹⁵ molecules of CO₂ make up -

Number of CO₂ moles present - 2.0756 x 10⁻⁹ mol of CO₂
For this problem, we should use the Henry's Law formula which is written below:
P = kC
where
P is the partial pressure of the gas
k is the Henry's Law constant at a certain temperature
C is the concentration
Substituting the values,
1.71 atm = (7.9×10⁻⁴<span> /atm)C
Solving for C,
C = 2164.56 molal or 2164.56 mol/kgwater
Let's make use of density of water (</span>1 kg/1 m³) and the molar mass of NF₃ (71 g/mol).<span>
Mass of NF</span>₃ = 2164.56 mol/kg water * 1 kg/1 m³ * 1 m³/1000000 mL * 150 mL * 71 g/mol = 23.05 g