What volume of ethylene glycol, a nonelectrolyte, must be added to 20.0 L of water to produce an antifreeze solution with a freezing point of −34.0 °C?
To calculate the volume of ethylene glycol needed, we can use the formula for freezing point depression:
\(\Delta T_f = i \cdot K_f \cdot m\)
Where:
\(\Delta T_f\) = freezing point depression
\(i\) = van't Hoff factor (1 for nonelectrolytes)
\(K_f\) = molal freezing point depression constant
\(m\) = molality of the solution
Since we want the antifreeze solution to have a freezing point of -34.0 °C, the freezing point depression (\(\Delta T_f\)) is:
\(\Delta T_f = 0 - (-34.0) = 34.0 °C\)
The molal freezing point depression constant for water is 1.86 °C/m.
The molality (\(m\)) can be calculated using the formula:
\(m = \frac{{moles\ of\ solute}}{{kg\ of\ solvent}}\)
We want to find the volume of ethylene glycol needed, so we need to convert the volume of water to kg.
The density of water is 1.00 g/mL, so the mass of 20.0 L (20,000 mL) of water is:
\(mass\ of\ water = (20,000\ mL) \times (1.00\ g/mL) = 20,000\ g = 20\ kg\)
Since ethylene glycol is a nonelectrolyte, we do not need to consider its dissociation in water.
Now, we can calculate the molality:
\(m = \frac{{moles\ of\ ethylene\ glycol}}{{20\ kg + mass\ of\ ethylene\ glycol}}\)
To find the moles of ethylene glycol needed to produce a solution with the desired freezing point depression, we can rearrange the formula and solve for moles of solute:
\(moles\ of\ ethylene\ glycol = m \times (20\ kg + mass\ of\ ethylene\ glycol)\)
Substituting back into our molality formula, we get:
\(m = \frac{{moles\ of\ ethylene\ glycol}}{{20\ kg + mass\ of\ ethylene\ glycol}}\)
Now we can plug in the values into the equation:
\(34.0\ °C = 1 \times 1.86\ °C/m \times \frac{{moles\ of\ ethylene\ glycol}}{{20\ kg + mass\ of\ ethylene\ glycol}}\)
Rearranging the formula to solve for moles of ethylene glycol:
\[moles\ of\ ethylene\ glycol = \left(\frac{{34.0\ °C}}{{1 \times 1.86\ °C/m}}\right) \times \left(20\ kg + mass\ of\ ethylene\ glycol\right)\]
Simplifying the equation:
\[moles\ of\ ethylene\ glycol = \left(\frac{{34.0\ °C}}{{1 \times 1.86\ °C/m}}\right) \times 20\ kg + \left(\frac{{34.0\ °C}}{{1 \times 1.86\ °C/m}}\right) \times mass\ of\ ethylene\ glycol\]
Since ethylene glycol is a liquid, we can approximate its density as 1.11 g/mL.
Converting the mass of ethylene glycol to kg, we have:
\[mass\ of\ ethylene\ glycol = (volume\ of\ ethylene\ glycol) \times (1.11\ g/mL) = volume\ of\ ethylene\ glycol\ (mL) \times (1.11\ g/mL)\]
\[mass\ of\ ethylene\ glycol = volume\ of\ ethylene\ glycol\ (mL) \times \left(\frac{{1.11\ g}}{{1\ mL}}\right) \times \left(\frac{{1\ kg}}{{1000\ g}}\right) = \frac{{volume\ of\ ethylene\ glycol}}{{1000}}\ kg\]
Substituting the mass conversion into the equation, we have:
\[moles\ of\ ethylene\ glycol = \left(\frac{{34.0\ °C}}{{1 \times 1.86\ °C/m}}\right) \times 20\ kg + \left(\frac{{34.0\ °C}}{{1 \times 1.86\ °C/m}}\right) \times \left(\frac{{volume\ of\ ethylene\ glycol}}{{1000}}\ kg\right) \]
Now we can solve for the volume of ethylene glycol needed by rearranging the equation:
\[volume\ of\ ethylene\ glycol = \left(\frac{{moles\ of\ ethylene\ glycol}}{{34.0\ °C/(1 \times 1.86\ °C/m)}}\right) \times 1000\ kg - 20\ kg\]
Substituting the freezing point depression (\(\Delta T_f\)) and molal freezing point depression constant (\(K_f\)):
\[volume\ of\ ethylene\ glycol = \left(\frac{{moles\ of\ ethylene\ glycol}}{{34.0\ °C/(1 \times 1.86\ °C/m)}}\right) \times 1000\ kg - 20\ kg\]
Finally, solve for the volume:
\[volume\ of\ ethylene\ glycol = \left(\frac{{moles\ of\ ethylene\ glycol}}{{34.0\ °C/(1 \times 1.86\ °C/m)}}\right) \times 1000\ kg - 20\ kg\]
Please note that the molar mass of ethylene glycol is 62.07 g/mol.