can someone please help me with this? i've been stuck on this one forever...
A solution is made by mixing 25.0 mL of toluene C6H5CH3d=0867gmL with 135.0 mL of benzene C6H6d=0874gmL. Assuming that the volumes add upon mixing, the molarity (M) and molality (m) of the toluene are
M and m.
see above.
To find the molarity (M) and molality (m) of toluene in the given solution, we need to use the formulas:
1. Molarity (M): Molarity is defined as the number of moles of solute per liter of solution (mol/L). The formula is:
M = (moles of solute) / (volume of solution in liters)
2. Molality (m): Molality is defined as the number of moles of solute per kilogram of solvent (mol/kg). The formula is:
m = (moles of solute) / (mass of solvent in kg)
First, let's calculate the moles of toluene:
The molecular weight of toluene (C6H5CH3) = (6 × 12.01 g/mol) + (6 × 1.01 g/mol) + 1 × 12.01 g/mol + 3 × 1.01 g/mol = 92.14 g/mol
Mass of toluene = volume × density = 25.0 mL × 0.867 g/mL = 21.675 g
Moles of toluene = mass / molecular weight = 21.675 g / 92.14 g/mol ≈ 0.235 mol
Now, let's calculate the molarity (M) of toluene:
Volume of solution = volume of toluene + volume of benzene = 25.0 mL + 135.0 mL = 160.0 mL = 0.160 L
Molarity (M) = moles of toluene / volume of solution in L = 0.235 mol / 0.160 L ≈ 1.47 M
Finally, let's calculate the molality (m) of toluene:
Mass of solvent (benzene) = volume × density = 135.0 mL × 0.874 g/mL = 118.05 g
Molality (m) = moles of toluene / mass of benzene in kg = 0.235 mol / 0.11805 kg ≈ 1.99 mol/kg
Therefore, the molarity (M) of toluene in the solution is approximately 1.47 M, and the molality (m) of toluene is approximately 1.99 mol/kg.