i am stuck on this problem from school
A solution is made by mixing 30.0 mL of toluene C6H5CH3d=0.867gmL with 130.0 mL of benzene C6H6d=0.874gmL . Assuming that the volumes add upon mixing, the molarity (M) and molality (m) of the toluene are
How many grams of toluene do you have? How many moles is that?
What is the mass of the benzene?
Molarity=molesToulene/volumesolutioninLiters
molality=molesToulene/kgOfSolvent
To find the molarity (M) and molality (m) of the toluene solution, we'll need to calculate the number of moles of toluene first.
Let's start with finding the moles of toluene (C6H5CH3):
1. Calculate the mass of toluene:
mass = volume * density
mass = 30.0 mL * 0.867 g/mL = 26.010 g
2. Calculate the moles of toluene:
moles = mass / molar mass
The molar mass of toluene (C6H5CH3) is:
C = 12.01 g/mol
H = 1.008 g/mol
Adding them up: 12.01 g/mol + (5 * 1.008 g/mol) + 12.01 g/mol = 92.14 g/mol
moles = 26.010 g / 92.14 g/mol = 0.2823 mol
Now that we have the mole value, we can find the molarity (M) and molality (m):
Molarity (M) = moles / volume in liters:
volume = 30.0 mL + 130.0 mL = 160.0 mL = 0.1600 L
Molarity = 0.2823 mol / 0.1600 L = 1.7643 M
Molality (m) = moles / mass in kilograms:
mass = 26.010 g
mass in kilograms = mass / 1000
mass in kilograms = 26.010 g / 1000 = 0.02601 kg
Molality = 0.2823 mol / 0.02601 kg = 10.835 m
So, the molarity (M) of the toluene solution is 1.7643 M, and the molality (m) of the toluene is 10.835 m.