# A solution of 6.4g of an unknown compound in 100g of benzene boils at 81.7 degrees Celsius. Determine molar mass of the unknown?

### 3 Answers

- pisgahchemistLv 710 months ago
Boiling point of benzene = 80.1C, Kb of benzene = 2.53 C/m

Two answers which don't agree. Here we write one equation to solve for the molar mass of the solute in terms of the other variables (mass of solute, mass of solvent, increase in boiling point).

i=1 since nothing ionizes

m = molality = moles of solute / kg of solvent = n / kg

moles = mass / molar mass.... n = m / M ...... n = mass / kg / molar mass

ΔT = (i)(Kb)(molality)

ΔT = 2.52 C/(mol/kg) x (mass / kg / molar mass)

molar mass = 2.52 C/(mol/kg) x (mass / kg / ΔT)

molar mass = 2.52 C/(mol/kg) x (6.4g / 0.100 kg / 1.6C)

molar mass = 101 g/mol

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- Roger the MoleLv 710 months ago
(81.7 - 80.1)°C / (2.53°C/m) = 0.632411 m

(0.632411 mol/kg) x (0.100 kg) = 0.0632411 mol

(6.4 g) / (0.0632411 mol) = 101 g/mol

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- 10 months ago
Elevation of the boiling point of benzene(ΔTb) = [(81.7+273) - (80.2+273)]K = 1.5 K

If the molar mass of the unknown compound is M then the molal concentration of 6.4gm of this samples is,

m = (6.4× 1000)/(100 × M = 64/M.

We know that ΔTb = kb × m

or,1.5 = 2.65 × (64/M)

or, M = (2.65×64)/1.5

∴ M = 113 gm/mol

Source(s): https://www.priyamstudycentre.com/- pisgahchemistLv 710 months agoReport
No. This answer is off the mark, due to differences in the boiling point elevation constant. 2.52 C/m vs 2.65 C/m used by Kajal, and the slight difference between the boiling point and the given temperature. Otherwise, this would give the same results as mine and Roger's.

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