Question

What is the boiling point in degrees Celsius of a solution formed by dissolving 3.75 mol of KCl in 1.00 kg of water?It is understood that the unit is ∘𝐶. Do not include units. Only enter the numerical value

Answer 1

Boiling point evaluation =$\Delta T_b=K_b×m$

Kb for water =0.51°c/m

$\Delta T_b=(0.51°c/m×3.75)$

$\Delta T_b=1.9125°C$

Hence the new boiling point is 101.9°C