Calculate the molar mass of a compound if 15.0 g dissolved in 70 g of ethanol

Question

(kb = 1.22 oC/m) caused the boiling point of ethanol to increase from 78.5 oC
to 84 oC.

Answer 1

To calculate the molar mass of the compound, we need to use the formula:

ΔT = (m * Kb) / (molar mass * molality)

Where:
ΔT = change in boiling point
m = mass of compound dissolved in solvent (ethanol)
Kb = boiling point elevation constant
molality = moles of solute / mass of solvent (ethanol)

First, let's find the molality of the solution:

Mass of solvent (ethanol) = 70 g
Molar mass of ethanol (C2H5OH) = 46.07 g/mol
Moles of solvent (ethanol) = mass of solvent / molar mass of ethanol = 70 g / 46.07 g/mol = 1.52 mol

molality = moles of solute / mass of solvent (ethanol)
molality = (15.0 g / molar mass) / 70 g

Now, let's calculate ΔT:

ΔT = change in boiling point = boiling point after - boiling point before = 84 °C - 78.5 °C = 5.5 °C

We can rearrange the formula to solve for the molar mass:

molar mass = (m * Kb) / (ΔT * molality)

Substituting the given values:

molar mass = (15.0 g * 1.22 °C/m) / (5.5 °C * (15.0 g / molar mass) / 70 g)

To simplify the expression, let's cancel out the units:

molar mass = (15.0 * 1.22) / (5.5 * (15.0 / molar mass) / 70)
molar mass = 18.3 / (5.5 * (1.0 / molar mass) / 70)

Let's simplify further by multiplying the fractions:

molar mass = 18.3 * (70 / 5.5) * molar mass / 1.0
molar mass = 18.3 * (70 / 5.5)

molar mass = 234.5454 g/mol

Therefore, the molar mass of the compound is approximately 234.55 g/mol.