Question

Ethylene glycol (CH₂OHCH₂OH) is used in antifreeze because, when mixed with water, it lowers the freezing point below $\mathbf{0}\mathbf{°}\mathbf{C}$. What mass percentage of ethylene glycol in water must be used to reduce the freezing point of the mixture to $\mathbf{-}\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{°}\mathbf{C}$, assuming ideal solution behavior?

Short answer

The Mass percentage of ethylene glycol is 14.35% .

Step-by-step solution

Given data

Ethylene glycol (CH₂OHCH₂OH) is used in antifreeze because, when mixed with water, it lowers the freezing point below$0°\mathrm{C}$

Concept  of Freezing point

Freezing point, temperature at which a liquid becomes a solid. As with the melting point, increased pressure usually raises the freezing point. The freezing point is lower than the melting point in the case of mixtures and for certain organic compounds such as fats.

Calculation of the mass of solution 

As we know that:

$∆\mathrm{T}$_f=K_fm

$∆\mathrm{T}$_f= T_{sham t} $-$T_station

$∆\mathrm{T}$_f= $0.0°\mathrm{C}-(-5.0°\mathrm{C})$

$∆\mathrm{T}$_f=$5.0°\mathrm{C}$

$∆\mathrm{T}$_f= 5.0K

$\mathrm{Since},5.0\mathrm{K}=1.86\mathrm{Kkgmol}$^-1$\times \mathrm{m}$

Therefore,

$\mathrm{m}=\frac{5.0\mathrm{K}}{1.86{\mathrm{Kkgmol}}^{-1}}$

$$\begin{gathered} \mathrm{m}=2.7\mathrm{mol}/\mathrm{K} \\ \mathrm{mass}=2.7\mathrm{mol}\times 62.068\mathrm{g}/\mathrm{mol} \\ \mathrm{mass}=167.58\mathrm{g} \end{gathered}$$

Calculation of the percentage mass

The mass of the solution is the sum of the masses.

167.58g + 1000g = 1167.58g.

$$\begin{gathered} \mathrm{Percentile}\mathrm{mass}=\frac{\mathrm{mass}\mathrm{of}\mathrm{solute}}{\mathrm{mass}\mathrm{of}\mathrm{solution}}\times 100\% \\ \mathrm{Percentile}\mathrm{mass}=\frac{167.58\mathrm{g}}{1,167.58\mathrm{g}}\times 100\% \\ \mathrm{Percentile}\mathrm{mass}=14.35\% \end{gathered}$$

Thus, the mass percentage of ethylene glycol is 14.35%.