Option 1
Idea of mole fraction: Instead of measuring how much of a substance is present by mass, mole fraction measures it by counting moles. If a solution is built from several substances, the mole fraction of any one substance is its share of the total moles, \(x_i = n_i / n_{total}\), and adding up every mole fraction gives exactly 1. It is a pure number and does not change with temperature.
Working the problem:
Step 1: Take a convenient 100 g sample of solution. 20% by mass glycol puts 20 g glycol and 80 g water in that sample.
Step 2: Convert each mass to moles using molar mass = mass / molar mass. Glycol (\(C_2H_6O_2\), 62 g/mol): \(20/62 = 0.323\) mol. Water (18 g/mol): \(80/18 = 4.44\) mol.
Step 3: Total moles \(= 0.323 + 4.44 = 4.77\) mol.
Step 4: Mole fraction of glycol \(= 0.323 / 4.77\).
\[ \boxed{x_{glycol} = 0.068,\quad x_{water} = 0.932} \]
As a check, \(0.068 + 0.932 = 1\), which confirms the answer.
Option 2
i) NaOH molarity:
Step 1: Convert 5 g of NaOH to moles. With a formula mass of 40 g/mol, moles \(= 5/40 = 0.125\) mol.
Step 2: Express the volume in litres: 450 mL \(= 0.45\) L.
Step 3: Divide moles by litres: \(0.125 / 0.45 = 0.278\).
\[ \boxed{0.278\ M} \]
ii) Acetic acid molality:
Step 1: Acetic acid, \(CH_3COOH\), has a molar mass of 60 g/mol, so 2.5 g gives \(2.5/60 = 0.0417\) mol.
Step 2: Molality uses the mass of solvent in kilograms: 75 g of benzene \(= 0.075\) kg.
Step 3: Divide: \(0.0417 / 0.075 = 0.556\).
\[ \boxed{0.556\ m} \]
Note that molality (unlike molarity) uses solvent mass, so it stays the same at any temperature.