Question

The molecular weight of an organic compound is 58 g mol⁻¹. What will be the boiling point of a solution containing 48 grams of the solute in 1200 grams of water? 
[K_b for water = 0.513°C kg mol⁻¹, Boiling point of water = 100°C]

Answer 1

M_B = 58 g mol⁻¹

T_b = ?

`T_b^0` = 100°C

W_B = 48 g

W_A = 1200 g

K_b = 0.513°C kg mol⁻¹

We know that

`Delta T_b = (K_b xx W_B xx 1000)/(W_A xx M_B)`

= `(0.513 xx 48 xx 1000)/(1200 xx 58)`

= 0.353°C

`T_b = T_b^0 + Delta T_b`

= 100 + 0.353

= 100.353°C