A solution was made by dissolving 3.75g of a solute in 208.7 g of acetone.the solution boiled at 56.58°c.the boiling point of oure acetone is 55.95°c,and the kb=1.71° cm/. What is the molecular weight of the solute
First, we need to calculate the molality (m) of the solution:
m = moles of solute / kg of solvent
Since acetone has a density of 0.786 g/mL, we can convert 208.7 g to liters:
V = 208.7 g / 0.786 g/mL = 265.39 mL = 0.26539 L
We can then calculate the moles of solute:
mol solute = 3.75 g / M
where M is the molecular weight of the solute.
Next, we need to convert the boiling point elevation (ΔTb) from Celsius to Kelvin:
ΔTb = Tb - Tb° = 56.58°C - 55.95°C = 0.63°C = 0.63 K
We can now use the following equation to solve for the molecular weight:
ΔTb = Kb x m
0.63 K = 1.71°C/m x m
m = 0.368 mol/kg
0.368 mol/kg = mol solute / 0.26539 kg
mol solute = 0.368 mol/kg x 0.26539 kg = 0.0978 mol
0.0978 mol = 3.75 g / M
M = 38.3 g/mol
Therefore, the molecular weight of the solute is 38.3 g/mol.