A solution is made by mixing 20.0 mL of toluene C6H5CH3d=0867gmL with 140.0 mL of benzene C6H6d=0874gmL. Assuming that the volumes add upon mixing, the molarity (M) and molality (m) of the toluene are?
There was a problem posted last night that used 150 mL benzene. Check to make sure that is not a typo.
Determine mass 20.0 mL toluene. Use density to do that. Then determine mols of toluene in that 20 mL.
Determine mass of benzene using density.
molality toluene = mols toluene/kg benzene.
Molarity toluene = mols/L .
To find the molarity (M) and molality (m) of toluene in the solution, we need to know the number of moles of toluene present.
First, let's calculate the moles of toluene:
Moles of toluene = volume of toluene × density of toluene ÷ molar mass of toluene
Given:
Volume of toluene = 20.0 mL
Density of toluene = 0.867 g/mL (density given in g/mL)
Molar mass of toluene = 92.14 g/mol
Moles of toluene = 20.0 mL × 0.867 g/mL ÷ 92.14 g/mol
Moles of toluene = 0.1864 mol
Now, let's find the molarity (M) of toluene:
Molarity (M) = moles of toluene ÷ volume of solution (in liters)
Given:
Volume of solution = volume of toluene + volume of benzene = 20.0 mL + 140.0 mL = 160.0 mL = 0.160 L
Molarity (M) = 0.1864 mol ÷ 0.160 L
Molarity (M) = 1.165 M
Finally, let's find the molality (m) of toluene:
Molality (m) = moles of toluene ÷ mass of the solvent (in kg)
Given:
Mass of the solvent = volume of benzene × density of benzene = 140.0 mL × 0.874 g/mL = 122.36 g = 0.12236 kg
Molality (m) = 0.1864 mol ÷ 0.12236 kg
Molality (m) ≈ 1.522 m
Therefore, the molarity (M) of toluene in the solution is approximately 1.165 M, and the molality (m) of toluene is approximately 1.522 m.