Question

1. to lower the freezing point of coolant in your automobile engine to 10° f, the solution of water and ethylene glycol, c₂h₆o₂, must be 65.0% antifreeze (volume to volume). what is the mole ratio of water to this toxic alcohol? (h₂o(l) d = 1.00 g/cm³) (c₂h₆o₂(l) d = 1.11 g/cm³)

Explanation:

Step1: Assume the volume of the solution

Let's assume the volume of the solution is \(V = 100\space cm^{3}\). Then the volume of ethylene glycol \(V_{1}=65\space cm^{3}\), and the volume of water \(V_{2}=(100 - 65)\space cm^{3}=35\space cm^{3}\).

Step2: Calculate the mass of each component

Using the formula \(m =
ho V\) (where \(m\) is mass, \(
ho\) is density, and \(V\) is volume).
For ethylene glycol: \(m_{1}=
ho_{1}V_{1}\), with \(
ho_{1} = 1.11\space g/cm^{3}\) and \(V_{1}=65\space cm^{3}\), so \(m_{1}=1.11\times65 = 72.15\space g\).
For water: \(m_{2}=
ho_{2}V_{2}\), with \(
ho_{2}=1.00\space g/cm^{3}\) and \(V_{2} = 35\space cm^{3}\), so \(m_{2}=1.00\times35=35\space g\).

Step3: Calculate the number of moles of each component

Using the formula \(n=\frac{m}{M}\) (where \(n\) is the number of moles, \(m\) is mass, and \(M\) is molar mass).
The molar mass of ethylene glycol \(M_{1}=(2\times12)+(6\times1)+(2\times16)=62\space g/mol\). So \(n_{1}=\frac{m_{1}}{M_{1}}=\frac{72.15}{62}\approx1.164\space mol\).
The molar mass of water \(M_{2}=(2\times1)+16 = 18\space g/mol\). So \(n_{2}=\frac{m_{2}}{M_{2}}=\frac{35}{18}\approx1.944\space mol\).

Step4: Calculate the mole ratio

The mole ratio of water to ethylene glycol is \(\frac{n_{2}}{n_{1}}=\frac{1.944}{1.164}\approx1.67\).

Answer:

The mole ratio of water to ethylene glycol is approximately \(1.67:1\).