Chapter 12 - Chemical Equilibrium - Exercise 12.1 - Concentrations of Pure Solids and Liquids - Page 526: b

Concentration of benzene: 11.3 mol/L

Molar mass of benzene $(C_6H_6)$ = 12.0 * 6 + 1.01 * 6 = 78.1 g/mol $$\frac{0.880 \space g}{1 \space mL} \times \frac{1000 \space mL}{1 \space L} \times \frac{1 \space mol}{78.1 \space g} = 11.3 \space mol/L$$