Question #e0398

Dec 8, 2014

The answer is $75.6 g r a m s$.

We are dealing with boiling-point elevation, which states that a solution has a higher boiling point than a pure solvent; that is, a solvent's boiling point (in this case, water's) will increase with the adition of, in this case, ethylene glycol.

The ecuation that models boiling-point elevation is

$\Delta {T}_{b o i l i n g} = {k}_{b} \cdot m$, where

$\Delta T$ - boiling point elevation (${T}_{f i n a l} - {T}_{i n i t i a l}$);
${k}_{b}$ - ebullioscopic constant;
$m$ - molality (NOT mass);

Water's ebullioscopic constant is usually given, at its normal boiling point being ${k}_{b}$ = $0.512 \frac{\mathrm{de} g r e e s C \cdot k g}{m o l e}$.

We know that $\Delta {T}_{b}$ = $102.5 - 100$ = ${2.5}^{\circ} C$, therefore we can determine the solution's molality

$m = \frac{\Delta {T}_{b}}{k} _ b = \frac{{2.5}^{\circ} C}{\frac{{0.512}^{\circ} C \cdot k g}{m o l e}} = 4.88 \frac{m o l e s}{k g}$. However, molality is defined as moles of solute devided by mass of solvent (in kg), which means

$m = {n}_{s o l u t e} / \left(m a s {s}_{s o l v e n t}\right) \to {n}_{s o l u t e} = m \cdot m a s {s}_{s o l u t e} = 4.88 \frac{m o l e s}{k g} \cdot 250 \cdot {10}^{- 3} k g = 1.22 m o l e s$

Knowing that ethylene glycol's molar mass is $62 \frac{g}{m o l e}$, we can get its mass by

$m a s s = n \cdot m o l a r m a s s = 1.22 m o l e s \cdot 62 \frac{g}{m o l e} = 75.6 g r a m s$.