**Part A**. Percent mass.

50.0 mL toluene (C_{6}H_{5}CH_{3}, *d* = 0.867 g/cm^{3})

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

125 mL benzene (C_{6}H_{6}, *d* = 0.874 g/cm^{3})

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

A solution is prepared by mixing 50.0 mL toluene (C_{6}H_{5}CH_{3}, *d* = 0.867 g/cm^{3}) with 125 mL benzene (C_{6}H_{6}, *d* = 0.874 g/cm^{3}). Assuming that the volumes add on mixing, calculate the mass percent, mole fraction, molality, and molarity of the toluene.

Frequently Asked Questions

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Making Solutions concept. If you need more Making Solutions practice, you can also practice Making Solutions practice problems.

What professor is this problem relevant for?

Based on our data, we think this problem is relevant for Professor Selampinar's class at UCONN.

What textbook is this problem found in?

Our data indicates that this problem or a close variation was asked in Chemistry: An Atoms First Approach - Zumdahl Atoms 1st 2nd Edition. You can also practice Chemistry: An Atoms First Approach - Zumdahl Atoms 1st 2nd Edition practice problems.