What would the final freezing point of water be if 3 mol of sugar were added to 1 kg of water ([tex]K_f=1.86^{\circ} C /( mol / kg )[/tex] for water and [tex]/=1[/tex] for sugar)?
A. [tex]+5.58^{\circ} C[/tex]
B. [tex]-1.86^{\circ} C[/tex]
C. [tex]-0.62^{\circ} C[/tex]
D. [tex]-5.58^{\circ} C[/tex]
Answer (2)
Calculate the molality of the solution: m = 1 kg 3 mol = 3 mol/kg .
Calculate the freezing point depression: Δ T f = 1 ⋅ 1.86 mol/kg ∘ C ⋅ 3 mol/kg = 5.5 8 ∘ C .
Calculate the new freezing point: T f , n e w = 0 ∘ C − 5.5 8 ∘ C = − 5.5 8 ∘ C .
The final freezing point of the water is − 5.5 8 ∘ C .
Explanation
Problem Analysis We are asked to find the final freezing point of water when 3 mol of sugar is added to 1 kg of water. We are given the freezing point depression constant for water, K f = 1.8 6 ∘ C / ( m o l / k g ) , and the van't Hoff factor for sugar, i = 1 .
Calculating Molality First, we need to calculate the molality of the solution. Molality is defined as the number of moles of solute per kilogram of solvent. In this case, the solute is sugar and the solvent is water. The number of moles of sugar is 3 mol, and the mass of water is 1 kg. Therefore, the molality of the solution is: m = m w a t er n s ug a r = 1 kg 3 mol = 3 mol/kg
Calculating Freezing Point Depression Next, we need to calculate the freezing point depression, Δ T f . The freezing point depression is given by the formula: Δ T f = i ⋅ K f ⋅ m where i is the van't Hoff factor, K f is the freezing point depression constant, and m is the molality. Plugging in the given values, we get: Δ T f = 1 ⋅ 1.86 mol/kg ∘ C ⋅ 3 mol/kg = 5.5 8 ∘ C
Calculating New Freezing Point Finally, we need to calculate the new freezing point of the solution. The new freezing point is given by: T f , n e w = T f , p u re − Δ T f where T f , p u re is the freezing point of pure water, which is 0 ∘ C . Plugging in the values, we get: T f , n e w = 0 ∘ C − 5.5 8 ∘ C = − 5.5 8 ∘ C
Final Answer Therefore, the final freezing point of the water is − 5.5 8 ∘ C .
Examples
The concept of freezing point depression is used in real life to de-ice roads during winter. Salt (NaCl) is added to the roads, which dissolves in the water (ice) and lowers the freezing point of the water. This prevents the water from freezing and forming ice, making the roads safer for driving. The amount of salt added needs to be calculated based on the expected temperature drop to ensure effective de-icing.
The final freezing point of water after adding 3 mol of sugar to 1 kg of water is -5.58 °C. This is calculated using the molality and the freezing point depression constant for water. The correct choice is D: -5.58 °C.
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