Question

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

Answer 1

Answer:

mass % = 28.4%

mole fraction = 0.252

molality = 3.08 molal

molarity = 2.69 M

Explanation:

Step 1: Data given

Volume of toluene = 50.0 mL = 0.05 L

Density toluene = 0.867 g/mL

Molar mass toluene = 92.14 g/mol

Volume of benzene = 125 mL = 0.125 L

Density benzene = 0.874 g/mL

Molar mass benzene = 78.11 g/mol

Step 2: Calculate masses

Mass = density * volume

Mass toluene = 50.0 mL * 0.867 g/mL = 43.35 g

Mass benzene = 125 mL * 0.874 g/mL = 109.25 g

Step 3: Calculate number of moles

Moles = mass / molar mass

Moles toluene = 43.35 grams /92.14 g/mol = 0.470 5 moles

Moles benzene = 109.25 grams / 78.11 g/mol = 1.399 moles

Step 4: Calculate molarity of toluene

Molarity = moles / volume

Molarity toluene = 0.4705 moles / 0.175 L = 2.69 M

Step 5: Calculate mass % of toluene

Mass % = (43.35 grams / (43.35 + 109.25) )*100 % = 28.4 %

Step 6: Calculate mole fraction of toluene

Mole fraction toluene = Moles toluene / total number of moles

Mole fraction toluene = 0.4705 / (0.4705 + 1.399) = 0.252

Step 7: Molality of toluene

Molality = number of moles / mass

Molality of toluene = 0.4705 moles / (0.04335 + 0.10925)

Molality of toluene = 3.08 molal