The boiling point of pure ethanol, c2h5oh, is 78.4 ∘c. its boiling point elevation constant is 1.22 °c/m. what is the boiling point of a solution formed by dissolving 8.00 g of alpha-naphthol (c10h7oh) in 100.0 g ethanol. 91.3 °c 79.1°c 97.6 °c 78.5°c

- B) 79.1 degree C

Solution:- An elevation in boiling point takes place when a non volatile solute is added to the solvent and the equation used for this is:

where, is the elevation in boiling point

i is Van't hoff factor. alpha naphthol is covalent compound and so the value of i is 1.

m is molality of the solution and kb is the molal elevation constant for the solvent and it's value is given as 1.22

Molar mass of alpha-naphthol = 10(12)+8(1)+1(16)

= 120 + 8 + 16

= 144 gram per mole

let's calculate the moles of solute:

= 0.0556 mole

molality is moles of solute per kg of solvent. 0.0556 moles of solute are dissolved in 100.0 g that is 0.100 kg of ethanol.

So, molality =

molality = 0.556m

Let's plug in the values in the formula and solve for boiling point elevation:

= 0.678

Boiling point of the solution = 78.4 + 0.678 = 79.078 degree C

If we round this two one decimal place then it becomes 79.1 degree C.  Hence, choice B) 79.1 degree C is the right answer.

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