The vapor pressure of benzene is 53.3kPa at 60.6∘ but it falls to 51.5kPa when 19 g of a non-volatile organic compound is dissolved in 500 g benzene. The molar mass of the non-volatile compound is close to:

Given Data

• Vapor pressure of pure benzene (P⁰_benzene): 53.3 kPa
• Vapor pressure of benzene solution (P_solution): 51.5 kPa
• Mass of non-volatile solute: 19 g
• Mass of benzene (solvent): 500 g
• Molar mass of benzene: 78 g/mol

1. Calculate Relative Lowering of Vapor Pressure

The relative lowering of vapor pressure is given by:

ΔP/P⁰_benzene = (P⁰_benzene - P_solution)/P⁰_benzene

Substituting the values:

ΔP/P⁰_benzene = (53.3 - 51.5)/53.3 = 0.0338

2. Determine the Moles of Benzene

Moles of benzene = Mass of benzene / Molar mass of benzene:

Moles = 500/78 = 6.41 mol

3. Apply Raoult's Law

According to Raoult's Law, for a dilute solution:

ΔP/P⁰_benzene ≈ Moles of solute / Moles of solvent

Rearranging:

Moles of solute = (ΔP/P⁰_benzene) * Moles of solvent

Substituting values:

Moles of solute = 0.0338 * 6.41 = 0.2166 mol

4. Calculate Molar Mass of Solute

Molar mass = Mass of solute / Moles of solute:

Molar mass = 19/0.2166 ≈ 87.7 g/mol

Final Answer

The molar mass of the non-volatile organic compound is closest to 88 g/mol, which corresponds to option 88.