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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 is approximately \(-18.166^{\circ}\mathrm{C}\).

Step by step solution

01

Determine the change in boiling point

First, we need to determine the change in boiling point of the aqueous solution. To do this, we will subtract the boiling point of pure water from the boiling point of the solution. \[ \Delta T_b = T_{solution} - T_{water} \] where \(T_{solution}\) = boiling point of the solution, which is given as 105.0°C, \(T_{water}\) = boiling point of pure water, which is 100°C, \[ \Delta T_b = 105.0 - 100 = 5.0^{\circ}\mathrm{C} \] So, the change in boiling point is 5.0°C.
02

Calculate the molality of the solute

Next, we need to determine the molality of the solute. To do this, we will use the boiling point elevation equation: \[ \Delta T_b = K_b \cdot m \] where \(\Delta T_b\) = change in boiling point, which is 5.0°C, \(K_b\) = molal boiling point elevation constant of water, which is 0.512°C/molal, \(m\) = molality of the solute. Solving for molality: \[ m = \frac{\Delta T_b}{K_b} = \frac{5.0}{0.512} \approx 9.766\, \mathrm{molal} \] So, the molality of the solute in the aqueous solution is approximately 9.766 molal.
03

Calculate the change in freezing point

Now, we need to determine the change in freezing point of the solution. To do this, we will use the freezing point depression equation: \[ \Delta T_f = K_f \cdot m \] where \(\Delta T_f\) = change in freezing point, \(K_f\) = molal freezing-point depression constant of water, which is 1.86°C/molal, \(m\) = molality of the solute, which is approximately 9.766 molal. Calculating \(\Delta T_f\): \[ \Delta T_f = 1.86 \cdot 9.766 \approx 18.166\,^{\circ}\mathrm{C} \] So, the change in freezing point is approximately 18.166°C.
04

Determine the freezing point of the solution

Finally, we need to determine the freezing point of the aqueous solution. To do this, we will subtract the change in freezing point from the freezing point of pure water. \[ T_{solution} = T_{water} - \Delta T_f \] where \(T_{solution}\) = freezing point of the solution, \(T_{water}\) = freezing point of pure water, which is 0°C, \(\Delta T_f\) = change in freezing point, which is approximately 18.166°C. Calculating the freezing point of the solution: \[ T_{solution} = 0 - 18.166 \approx -18.166\,^{\circ}\mathrm{C} \] So, the freezing point of the aqueous solution is approximately -18.166°C.

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Most popular questions from this chapter

Indicate whether each statement is true or false: (a) If you compare the solubility of a gas in water at two different temperatures, you find the gas is more soluble at the lower temperature. (b) The solubility of most ionic solids in water decreases as the temperature of the solution increases. (c) The solubility of most gases in water decreases as the temperature increases because water is breaking its hydrogen bonding to the gas molecules as the temperature is raised. (d) Some ionic solids become less soluble in water as the temperature is raised.

Some soft drinks contain up to 85 ppm oxygen. (a) What is this concentration in mol/L? (b) What partial pressure of \(\mathrm{O}_{2}\) above water is needed to obtain \(85 \mathrm{ppm} \mathrm{O}_{2}\) in water at $10^{\circ} \mathrm{C} ?\( (The Henry's law constant for \)\mathrm{O}_{2}$ at this temperature is $\left.1.69 \times 10^{-5} \mathrm{~mol} / \mathrm{m}^{3}-\mathrm{Pa} .\right)$

(a) What is the molality of a solution formed by dissolving 1.12 mol of KCl in 16.0 mol of water? (b) How many grams of sulfur \(\left(\mathrm{S}_{8}\right)\) must be dissolved in \(100.0 \mathrm{~g}\) of naphthalene $\left(\mathrm{C}_{10} \mathrm{H}_{8}\right)\( to make a \)0.12 \mathrm{~m}$ solution?

What is the molarity of each of the following solutions: (a) \(15.0 \mathrm{~g}\) of \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) in \(0.250 \mathrm{~mL}\) solution, (b) \(5.25 \mathrm{~g}\) of $\mathrm{Mn}\left(\mathrm{NO}_{3}\right)_{2} \cdot 2 \mathrm{H}_{2} \mathrm{O}\( in \)175 \mathrm{~mL}$ of solution, (c) \(35.0 \mathrm{~mL}\) of \(9.00 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) diluted to $0.500 \mathrm{~L} ?$

(a) Would you expect stearic acid, \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{16} \mathrm{COOH},\) to be more soluble in water or in carbon tetrachloride? (b) Which would you expect to be more soluble in water, cyclohexane or dioxane?
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