Question

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

Answer 1

Answer:- 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:

$\Delta T=i*m*kb$

where, $\Delta T$ 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:

$8.00g(\frac{1mole}{144g})$

= 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 = $\frac{0.0556mole}{0.1kg}$

molality = 0.556m

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

$\Delta T=(1)(0.556)(1.22)$

$\Delta T$ = 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.