Question

Calculate the boiling point of 3.42 m solution of ethylene glycol, C2H602, in water (Kb =0.51°C/m, the normal boiling point of water is 100°C):​

Answer 1

Answer:

128.10 degree Celsius

Explanation:

The boiling point is given by

$T_b = \frac{1000*K_b * w}{M*W}$

Substituting the given values, we get -

$T_b = \frac{1000*0.51 * 3.42}{62.07}\\T_b = 28.10$

Tb is the change in temperature

T2 -T1 = 28.10

T2 = 28.10 +100

T2 = 128.10 degree Celsius