# Question #d5d1d

Dec 15, 2014

The answer is ${0.56}^{\circ} C$.

Boiling-point elevation is modeled by this equation

$\Delta {T}_{b} = {k}_{b} \cdot m$, where

$\Delta {T}_{b}$ - 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 $\rho = 1.0 \frac{k g}{L}$, the mass of the solvent will be

${m}_{w a t e r} = {\rho}_{w a t e r} \cdot V = 1.0 \frac{k g}{L} \cdot 2.0 L = 2.0 k g$

So, the solution's molality wil be

$m = \frac{2.1 m o l e s}{2.0 k g} = 1.1 m$

Therefore, the boiling-point elevation will be

$\Delta {T}_{b} = {0.51}^{\circ} \frac{C}{m} \cdot 1.1 m = {0.56}^{\circ} C$