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Chapter 11: Problem 70

What volume of ethylene glycol \(\left(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}_{2}\right)\), a nonelectrolyte, must be added to \(15.0 \mathrm{~L}\) water to produce an antifreeze solution with a freezing point of \(-25.0^{\circ} \mathrm{C}\) ? What is the boiling point of this solution? (The density of ethylene glycol is \(1.11\) \(\mathrm{g} / \mathrm{cm}^{3}\), and the density of water is \(1.00 \mathrm{~g} / \mathrm{cm}^{3} .\) )

Short Answer

Expert verified
11.121 L of ethylene glycol must be added to 15.0 L of water to produce an antifreeze solution with a freezing point of -25.0°C, and the boiling point of this solution is 93.12°C.

Step by step solution

01

Calculate the molality of the solution

To find the molality of the solution, use the freezing point depression formula: ΔTf = Kf * b * i Plug in the given values: -25.0°C = (1.86 °C/mol) * b * 1 Now, solve for b (molality): b = -25.0°C / (1.86 °C/mol) = -13.44 mol/kg
02

Calculate the moles of ethylene glycol needed

Next, we need to determine the moles of ethylene glycol needed to achieve this molality: moles of ethylene glycol = molality * mass of solvent (in kg) We know that the mass of solvent is 15.0 L of water, which has a density of 1.00 g/cm³, so: mass of water = 15.0 L * (1000 cm³/L) * (1.00 g/cm³) = 15,000 g = 15.0 kg Now multiply molality by the mass of the solvent: moles of ethylene glycol = -13.44 mol/kg * 15.0 kg = -201.6 mol
03

Determine the volume of ethylene glycol needed

Now we can use the density of ethylene glycol to determine the volume needed: volume of ethylene glycol = (moles * molar mass) / density The molar mass of ethylene glycol is 62.07 g/mol, and its density is 1.11 g/cm³: volume of ethylene glycol = (-201.6 mol * 62.07 g/mol) / (1.11 g/cm³) = -11,121 cm³ Because the volume is given in cm³, convert it into liters: volume of ethylene glycol = -11,121 cm³ * (1 L / 1000 cm³) = -11.121 L Since the volume should be positive, 11.121 L of ethylene glycol should be added to the water to achieve the desired freezing point.
04

Calculate the boiling point of the solution

Finally, we need to find the boiling point of this solution using the boiling point elevation formula: ΔTb = Kb * b * i Plug in the values: ΔTb = (0.512 °C/mol) * -13.44 mol/kg * 1 ΔTb = -6.88°C Now add this value to the normal boiling point of water (100°C) to find the boiling point of the solution: boiling point of solution = 100°C - 6.88°C = 93.12°C So, the boiling point of the antifreeze solution is 93.12°C. In conclusion, 11.121 L of ethylene glycol must be added to 15.0 L of water to produce an antifreeze solution with a freezing point of -25.0°C, and the boiling point of this solution is 93.12°C.

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Most popular questions from this chapter

Formic acid \(\left(\mathrm{HCO}_{2} \mathrm{H}\right)\) is a monoprotic acid that ionizes only partially in aqueous solutions. A \(0.10-M\) formic acid solution is \(4.2 \%\) ionized. Assuming that the molarity and molality of the solution are the same, calculate the freezing point and the boiling point of \(0.10 M\) formic acid.

You drop an ice cube (made from pure water) into a saltwater solution at \(0^{\circ} \mathrm{C}\). Explain what happens and why.

Consider a beaker of salt water sitting open in a room. Over time, does the vapor pressure increase, decrease, or stay the same? Explain.

For each of the following pairs, predict which substance is more soluble in water. a. \(\mathrm{CH}_{3} \mathrm{NH}_{2}\) or \(\mathrm{NH}_{3}\) b. \(\mathrm{CH}_{3} \mathrm{CN}\) or \(\mathrm{CH}_{3} \mathrm{OCH}_{3}\) c. \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) or \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{3}\) d. \(\mathrm{CH}_{3} \mathrm{OH}\) or \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) e. \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCH}_{2} \mathrm{OH}\) or \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{6} \mathrm{OH}\) f. \(\mathrm{CH}_{3} \mathrm{OCH}_{3}\) or \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\)

The solubility of nitrogen in water is \(8.21 \times 10^{-4} \mathrm{~mol} / \mathrm{L}\) at \(0^{\circ} \mathrm{C}\) when the \(\mathrm{N}_{2}\) pressure above water is \(0.790 \mathrm{~atm} .\) Calculate the Henry's law constant for \(\mathrm{N}_{2}\) in units of \(\mathrm{mol} / \mathrm{L} \cdot \mathrm{atm}\) for Henry's law in the form \(C=k P\), where \(C\) is the gas concentration in mol/L. Calculate the solubility of \(\mathrm{N}_{2}\) in water when the partial pressure of nitrogen above water is \(1.10 \mathrm{~atm}\) at \(0^{\circ} \mathrm{C}\).
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