A solution is made by mixing 35.0 mL of toluene C6H5CH3 d=0867gmL with 125.0 mL of benzene C6H6 D=0874gmL. Assuming that the volumes add upon mixing, the molarity (M) and molality (m) of the toluene are?
Find M and m
molarity = # mols/L of solution.
molality = # mols/kg solvent.
g toluene = density x volume
g benzene = density x volume
mols benzene = g/molar mass
Post your work if you get stuck.
thanks i got it
To find the molarity (M) and molality (m) of the toluene solution, we first need to calculate the number of moles of toluene.
Step 1: Calculate the mass of toluene:
Mass of toluene = Volume × Density = 35.0 mL × 0.867 g/mL = 30.435 g
Step 2: Calculate the moles of toluene:
Moles of toluene = Mass ÷ Molar mass
The molar mass of toluene (C7H8) is:
(12.01 g/mol × 7) + (1.01 g/mol × 8) = 92.14 g/mol
Moles of toluene = 30.435 g ÷ 92.14 g/mol ≈ 0.33 mol
Step 3: Calculate the molarity (M) of toluene:
Molarity (M) = Moles ÷ Volume
Volume = 35.0 mL + 125.0 mL = 160.0 mL = 0.160 L (since 1 L = 1000 mL)
Molarity (M) = 0.33 mol ÷ 0.160 L ≈ 2.06 M
Step 4: Calculate the molality (m) of toluene:
Molality (m) = Moles ÷ Mass of the solvent (in kg)
Mass of the solvent (benzene) = Volume × Density = 125 mL × 0.874 g/mL = 109.25 g = 0.10925 kg
Molality (m) = 0.33 mol ÷ 0.10925 kg ≈ 3.02 m
Therefore, the molarity (M) of the toluene solution is approximately 2.06 M and the molality (m) of the toluene is approximately 3.02 m.