Problem: A solution is prepared by mixing 50.0 mL toluene (C6H5CH3, d = 0.867 g/cm3) with 125 mL benzene (C6H6, d = 0.874 g/cm3). Assuming that the volumes add on mixing, calculate the mass percent, mole fraction, molality, and molarity of the toluene.

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FREE Expert Solution

Part A. Percent mass.

50.0 mL toluene (C₆H₅CH₃, d = 0.867 g/cm³)

$\mathbf{50}\mathbf{.}\mathbf{0}\mathbf{ }\cancel{\mathbf{mL}}\mathbf{\times }\frac{\mathbf{1}\mathbf{ }\cancel{{\mathbf{cm}}^{\mathbf{3}}}}{\mathbf{1}\mathbf{ }\cancel{\mathbf{mL}}}\mathbf{\times }\frac{\mathbf{0}\mathbf{.}\mathbf{867}\mathbf{ }\mathbf{g}}{\cancel{{\mathbf{cm}}^{\mathbf{3}}}}\mathbf{=}$43.35 g

125 mL benzene (C₆H₆, d = 0.874 g/cm³)

$\mathbf{125}\mathbf{ }\cancel{\mathbf{mL}}\mathbf{\times }\frac{\mathbf{1}\mathbf{ }\cancel{{\mathbf{cm}}^{\mathbf{3}}}}{\mathbf{1}\mathbf{ }\cancel{\mathbf{mL}}}\mathbf{\times }\frac{\mathbf{0}\mathbf{.}\mathbf{847}\mathbf{ }\mathbf{g}}{\cancel{{\mathbf{cm}}^{\mathbf{3}}}}\mathbf{=}$105.875 g