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Chapter 13: Problem 76

What is the freezing point of an aqueous solution that boils at \(105.0^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
The freezing point of the aqueous solution that boils at \(105.0^{\circ}\mathrm{C}\) is approximately \(-18.17^{\circ}\mathrm{C}\).

Step by step solution

01

Determine the boiling point elevation of the solution.

The boiling point elevation, ΔTb, can be found by subtracting the normal boiling point of the solvent (water) from the boiling point of the solution: ΔTb = Tb - Tb,water In this case, the boiling point of water is \(100^{\circ}\mathrm{C}\). So: ΔTb = \(105.0^{\circ}\mathrm{C}\) - \(100^{\circ}\mathrm{C}\) = \(5^{\circ}\mathrm{C}\).
02

Find the molality of the solute in the solution.

The boiling point elevation formula is given by: ΔTb = Kb * molality Where Kb is the ebullioscopic constant, which for water is 0.512 \(^{\circ}\mathrm{C/m}\). Rearranging the equation and plugging in the values, we get: molality = ΔTb / Kb = \(\frac{5^{\circ}\mathrm{C}}{0.512^{\circ}\mathrm{C/m}}\) = 9.7656 m
03

Determine the freezing point depression of the solution.

The freezing point depression formula is given by: ΔTf = Kf * molality Where Kf is the cryoscopic constant, which for water is 1.86 \(^{\circ}\mathrm{C/m}\). Plugging in the values, we get: ΔTf = 1.86 \(^{\circ}\mathrm{C/m}\) * 9.7656 m = 18.1656 \(^{\circ}\mathrm{C}\)
04

Find the freezing point of the aqueous solution.

The freezing point of the solution can be found by subtracting the freezing point depression from the normal freezing point of the solvent (water). In this case, the freezing point of water is \(0^{\circ}\mathrm{C}\). So: Freezing point of the solution = Tf,water - ΔTf Freezing point of the solution = \(0^{\circ}\mathrm{C}\) - 18.1656 \(^{\circ}\mathrm{C}\) = \(-18.1656^{\circ}\mathrm{C}\) The freezing point of the aqueous solution that boils at \(105.0^{\circ}\mathrm{C}\) is approximately \(-18.17^{\circ}\mathrm{C}\).

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Most popular questions from this chapter

At \(35^{\circ} \mathrm{C}\) the vapor pressure of acetone, \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CO},\) is 360 torr, and that of chloroform, \(\mathrm{CHCl}_{3}\), is 300 torr. Acetone and chloroform can form very weak hydrogen bonds between one another as follows: A solution composed of an equal number of moles of acetone and chloroform has a vapor pressure of 250 torr at \(35^{\circ} \mathrm{C}\). (a) What would be the vapor pressure of the solution if it exhibited ideal behavior? (b) Use the existence of hydrogen bonds between acetone and chloroform molecules to explain the deviation from ideal behavior. (c) Based on the behavior of the solution, predict whether the mixing of acetone and chloroform is an exothermic \(\left(\Delta H_{\text {soln }}<0\right)\) or endothermic \(\left(\Delta H_{\text {soln }}>0\right)\) process.

Explain how (a) a soap such as sodium stearate stabilizes a colloidal dispersion of oil droplets in water; (b) milk curdles upon addition of an acid.

Calculate the molarity of the following aqueous solutions: (a) \(0.540 \mathrm{~g} \mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}\) in \(250.0 \mathrm{~mL}\) of solution, (b) \(22.4 \mathrm{~g}\) \(\mathrm{LiClO}_{4} \cdot 3 \mathrm{H}_{2} \mathrm{O}\) in \(125 \mathrm{~mL}\) of solution, (c) \(25.0 \mathrm{~mL}\) of \(3.50 \mathrm{M}\) \(\mathrm{HNO}_{3}\) diluted to \(0.250 \mathrm{~L}\)

At ordinary body temperature \(\left(37^{\circ} \mathrm{C}\right)\) the solubility of \(\mathrm{N}_{2}\) in water in contact with air at ordinary atmospheric pressure \((1.0 \mathrm{~atm})\) is \(0.015 \mathrm{~g} / \mathrm{L}\). Air is approximately \(78 \mathrm{~mol} \% \mathrm{~N}_{2}\). Calculate the number of moles of \(\mathrm{N}_{2}\) dissolved per liter of blood, which is essentially an aqueous solution. At a depth of \(100 \mathrm{ft}\) in water, the pressure is \(4.0 \mathrm{~atm}\). What is the solubility of \(\mathrm{N}_{2}\) from air in blood at this pressure? If a scuba diver suddenly surfaces from this depth, how many milliliters of \(\mathrm{N}_{2}\) gas, in the form of tiny bubbles, are released into the bloodstream from each liter of blood?

A "canned heat" product used to warm chafing dishes consists of a homogeneous mixture of ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) and paraffin that has an average formula of \(\mathrm{C}_{24} \mathrm{H}_{50}\). What mass of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) should be added to \(620 \mathrm{~kg}\) of the paraffin in formulating the mixture if the vapor pressure of ethanol at \(35^{\circ} \mathrm{C}\) over the mixture is to be 8 torr? The vapor pressure of pure ethanol at \(35^{\circ} \mathrm{C}\) is 100 torr.
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