A solution was made by dissolving 3.75g of a solite i 108.7g acetone.the solution boiled at 56.58°c.the boiling point of pure acetone is 55.95°c,and the kb=1.7+°c/m what is the molecular weight of the solute

Answer 1

First, we need to calculate the change in boiling point (ΔTb) of the solution:

ΔTb = Tb(solution) - Tb(pure acetone)
ΔTb = 56.58°C - 55.95°C
ΔTb = 0.63°C

Next, we can use the formula for the molality (m) of the solution:

ΔTb = kb * m
m = ΔTb / kb
m = 0.37 mol/kg

Now we can use the formula for molality:

m = moles of solute / mass of solvent (in kg)
0.37 = moles / 0.1087 kg

Solving for moles:

moles = m * mass of solvent
moles = 0.37 * 0.1087
moles = 0.04019 mol

Finally, we can use the formula for the molecular weight (M) of the solute:

M = molar mass / number of particles
M = (3.75 g / 0.04019 mol) / 1
M = 93.29 g/mol

Therefore, the molecular weight of the solute is 93.29 g/mol.