Question

a 50/50 blend of engine coolant and water (by volume) is usually used in an automobiles engine cooling system. if a cars cooling system holds 5.30 gal, what is the boiling point of the solution? for the calculation, assume that at normal filling conditions, the densities of engine coolant and water are 1.11 g/ml and 0.998 g/ml respectively. also, assume that the engine coolant is pure ethylene glycol (hoch₂ch₂oh), which is non - ionizing and non - volatile, and that the pressure remains constant at 1.00 atm. the boiling - point elevation constant for water will also be needed.

Explanation:

Step1: Calculate the volume of ethylene - glycol and water

The cooling system holds $V = 5.30$ gal. Since it is a 50/50 blend by volume, the volume of ethylene - glycol $V_{1}$ and water $V_{2}$ is $V_{1}=V_{2}=\frac{5.30}{2}=2.65$ gal. Convert gallons to milliliters: $1$ gal = $3785.41$ mL, so $V_{1}=V_{2}=2.65\times3785.41 = 10031.3465$ mL.

Step2: Calculate the mass of ethylene - glycol and water

Use the density formula $m=
ho V$. For ethylene - glycol with $
ho_{1}=1.11$ g/mL, $m_{1}=
ho_{1}V_{1}=1.11$ g/mL$\times10031.3465$ mL = $11134.794615$ g. For water with $
ho_{2}=0.998$ g/mL, $m_{2}=
ho_{2}V_{2}=0.998$ g/mL$\times10031.3465$ mL = $9991.283807$ g.

Step3: Calculate the molality of the solution

The molar mass of ethylene - glycol ($HOCH_{2}CH_{2}OH$) is $M=(2\times12.01 + 6\times1.01+2\times16.00)$ g/mol = $62.07$ g/mol. The number of moles of ethylene - glycol $n=\frac{m_{1}}{M}=\frac{11134.794615}{62.07}$ mol = $179.391$ mol. The mass of water in kg is $m_{2}=9991.283807$ g = $9.991283807$ kg. The molality $m=\frac{n}{m_{2}}=\frac{179.391}{9.991283807}$ mol/kg = $17.95$ mol/kg.

Step4: Calculate the boiling - point elevation

The boiling - point elevation constant for water $K_{b}=0.512^{\circ}C/m$. The boiling - point elevation $\Delta T_{b}=K_{b}m$. So $\Delta T_{b}=0.512^{\circ}C/m\times17.95$ m = $9.19$ $^{\circ}C$.

Step5: Calculate the boiling point of the solution

The normal boiling point of water at 1.00 atm is $T_{0}=100^{\circ}C$. The boiling point of the solution $T = T_{0}+\Delta T_{b}=100^{\circ}C + 9.19^{\circ}C=109.19^{\circ}C$.

Answer:

$109.19^{\circ}C$