When 0.270 mol of a nondissociating solute is dissolved in 410.0 ml of cs2, the solution boils at 47.52 ∘c. what is the molal boiling-point-elevation constant for cs2 ?

Kb = 0.428 m/°C

Explanation:

To solve this problem we need to use the boiling-point elevation formula:

Where Tsolution and Tpure solvent are the boiling point of the CS₂ solution (47.52 °C) and of pure CS₂ (46.3 °C), respectively. Kb is the constant asked by the problem, and m is the molality of the solution.

So in order to use that equation and solve for Kb, first we calculate the molality of the solution.

molality = mol solute / kg solvent

molality = 0.270 mol / 0.5166 kg = 0.5226 m

Now we solve for Kb:

Tsolution - Tpure solvent = Kb * m

b) nuclear ive learned that in a pedtrodic table