Question

ethylene glycol is used in the antifreeze you put in your car to lower the freezing point and raise the boiling point of water in the engine block - making sure it stays in liquid form. what mass of ethylene glycol (g) must be added to 1.0kg water to produce a solution that boils at 105.0°c?

Explanation:

Step1: Calculate the boiling - point elevation

The boiling - point elevation formula is \(\Delta T_b = K_b\times m\).
The boiling point of pure water \(T_{b,0}=100.0^{\circ}C\), and the boiling point of the solution \(T_b = 105.0^{\circ}C\). So \(\Delta T_b=T_b - T_{b,0}=105.0 - 100.0=5.0^{\circ}C\).
For water, \(K_b = 0.512^{\circ}C\cdot kg/mol\).
From \(\Delta T_b = K_b\times m\), we can solve for the molality \(m\): \(m=\frac{\Delta T_b}{K_b}\).
Substitute \(\Delta T_b = 5.0^{\circ}C\) and \(K_b = 0.512^{\circ}C\cdot kg/mol\) into the formula: \(m=\frac{5.0}{0.512}\approx9.77mol/kg\).

Step2: Calculate the moles of ethylene glycol

Molality \(m=\frac{n_{\text{ethylene glycol}}}{m_{\text{water (in kg)}}}\).
Given \(m_{\text{water}} = 1.0kg\) and \(m = 9.77mol/kg\).
Since \(m=\frac{n_{\text{ethylene glycol}}}{m_{\text{water (in kg)}}}\), then \(n_{\text{ethylene glycol}}=m\times m_{\text{water (in kg)}}\).
Substitute \(m = 9.77mol/kg\) and \(m_{\text{water}} = 1.0kg\) into the formula: \(n_{\text{ethylene glycol}}=9.77mol\).

Step3: Calculate the mass of ethylene glycol

The molar mass of ethylene glycol \((C_2H_6O_2)\) is \(M=(2\times12.01 + 6\times1.008+2\times16.00)g/mol=(24.02 + 6.048 + 32.00)g/mol = 62.068g/mol\).
The mass formula is \(m = n\times M\).
Substitute \(n = 9.77mol\) and \(M = 62.068g/mol\) into the formula: \(m=9.77\times62.068\approx606g\).

Answer:

\(606g\)