Question

Question #d5d1d

Answer 1

The answer is `0.56^@C`.

Boiling-point elevation is modeled by this equation

`DeltaT_b = k_b * m`, where

`DeltaT_b` - represents the boiling-point elevation;
`k_b` - represents water's ebullioscopic constant;
`m` - represents the solution's molality;

SInce we are dealing with water, which has a density of `rho = 1.0(kg)/L`, the mass of the solvent will be

`m_(water) = rho_(water) * V = 1.0 (kg)/L * 2.0L = 2.0 kg`

So, the solution's molality wil be

`m = (2.1 mol es)/(2.0 kg) = 1.1m`

Therefore, the boiling-point elevation will be

`DeltaT_b = 0.51^@C/m * 1.1m = 0.56^@C`