Question

Explain mole fraction. Calculate the mole fraction of ethylene glycol if 20% of the mass of ethylene glycol is present in the solution. [1+4]
OR
i) Calculate the molarity of the solution in which 5 g NaOH is dissolved in 450 mL of solution.
ii) Calculate the molality of the solution in which 2.5 g acetic acid is dissolved in 75 g benzene. [2½+2½]

Hint:

Mole fraction = moles of component divided by total moles (sum equals 1). For 20% glycol take 20 g glycol (M = 62) and 80 g water (M = 18). Molarity = moles of solute per litre of solution (NaOH M = 40, volume in L); molality = moles of solute per kilogram of solvent (acetic acid M = 60, benzene mass in kg).

Answer 1

Option 1

Mole fraction: The mole fraction of a component in a solution is the ratio of the number of moles of that component to the total number of moles of all components present. For component A in a mixture of A and B, \( x_A = \dfrac{n_A}{n_A + n_B} \). The sum of the mole fractions of all components is always 1.

Numerical: 20% by mass of ethylene glycol means 20 g of ethylene glycol are present in 100 g of solution, so the mass of water (solvent) = 100 − 20 = 80 g.
Step 1 (molar masses): Ethylene glycol is C₂H₆O₂ (HOCH₂CH₂OH), molar mass = (2×12) + (6×1) + (2×16) = 62 g mol^−1; water = 18 g mol^−1.
Step 2 (moles of glycol): \( n_{glycol} = \dfrac{20}{62} = 0.3226 \) mol.
Step 3 (moles of water): \( n_{water} = \dfrac{80}{18} = 4.444 \) mol.
Step 4 (mole fraction): \( x_{glycol} = \dfrac{0.3226}{0.3226 + 4.444} = \dfrac{0.3226}{4.767} \).
\[ \boxed{x_{glycol} \approx 0.068} \]
Hence the mole fraction of water = 1 − 0.068 = 0.932.

Option 2

i) Molarity of NaOH solution:
Molarity = number of moles of solute per litre of solution, \( M = \dfrac{n}{V(L)} \).
Step 1: Molar mass of NaOH = 23 + 16 + 1 = 40 g mol^−1.
Step 2: Moles of NaOH \( = \dfrac{5}{40} = 0.125 \) mol.
Step 3: Volume = 450 mL = 0.450 L.
Step 4: \( M = \dfrac{0.125}{0.450} \).
\[ \boxed{M \approx 0.278\ \text{mol L}^{-1}} \]

ii) Molality of acetic acid in benzene:
Molality = number of moles of solute per kilogram of solvent, \( m = \dfrac{n}{W_{solvent}(kg)} \).
Step 1: Molar mass of acetic acid CH₃COOH = (2×12) + (4×1) + (2×16) = 60 g mol^−1.
Step 2: Moles of acetic acid \( = \dfrac{2.5}{60} = 0.0417 \) mol.
Step 3: Mass of benzene = 75 g = 0.075 kg.
Step 4: \( m = \dfrac{0.0417}{0.075} \).
\[ \boxed{m \approx 0.556\ \text{mol kg}^{-1}} \]