Question

Acenapthalene has the empirical formula C6H5. A solution of 0.515 g of acenapthalene in 15.0 g CHCl3 boils at 62.5oC. The normal boiling point of CHCl3 is 61.7oC; the Kb is 3.63 oC/molal. What is the molecular formula of acenapthalene? (Atomic weights: C = 12.01, H = 1.008, Cl = 35.45).

Answer 1

The molecular formula of an ascenapthalene is
`C_(12)H_(10)`

Step-by-step explanation:

`\Delta T_b=K_b* m`

`\Delta T_b=K_b* \frac{\text{Mass of acenapthalene}}{\text{Molar mass of acenapthalene}* \text{Mass of chloroform in Kg}}`

where,

`\Delta T_f` =Elevation in boiling point =
`(62.5-61.7)^oC=0.8^oC`

Mass of acenapthalene = 0.515 g

Mass of
`CHCl_3` = 15.0 g = 0.015 kg (1 kg = 1000 g)

`K_b` = boiling point constant = 3.63 °C/m

m = molality

Now put all the given values in this formula, we get

`0.8^0C=3.67 ^oC/m* \frac{0.515}{\text{Molar mass of acenapthalene}* 0.015kg}`

`\text{Molar mass of acenapthalene}=155.7875 g/mol`

Let the molecule formula of the Acenapthalene be
`C_{6n]H_(5n)`

`6n* 12 g/mol+5n* 1 g/mol=155.7875 g/mol`

n = 2.0

The molecular formula of an ascenapthalene is
`C_(12)H_(10)`