What volume of ethanol (density $=0.7893 g / cm ^3$ ) should be added to 450. mL of water in order to have a solution that freezes at $-15.0^{\circ} C$ ? Assume the density of water is $1.0 g / mL$.
(For water, $K_f=1.86^{\circ} C / m$.)
A. 470 mL
B. 167 mL
C. 371 mL
D. 212 mL
E. 132 mL
Answer (1)
Calculate the mass of water: ma s s w a t er = 450 m L × 1.0 g / m L = 450 g = 0.45 k g .
Determine the freezing point depression: Δ T f = 0 ∘ C − ( − 1 5 ∘ C ) = 1 5 ∘ C .
Calculate the molality of the solution: m = K f Δ T f = 1.86 15 ≈ 8.0645 m .
Find the volume of ethanol: m o l e s e t han o l = 8.0645 × 0.45 ≈ 3.629 m o l , ma s s e t han o l = 3.629 × 46.068 ≈ 167.2 g , v o l u m e e t han o l = 0.7893 167.2 ≈ 211.8 m L . Thus, the final answer is 212 m L .
Explanation
Problem Setup and Given Information We are given the density of ethanol ( 0.7893 g / c m 3 ), the volume of water (450 mL), the density of water ( 1.0 g / m L ), the freezing point depression ( − 15. 0 ∘ C ), and the freezing point depression constant for water ( K f = 1.8 6 ∘ C / m ). Our goal is to find the volume of ethanol that needs to be added to the water to achieve the specified freezing point depression.
Calculating the Mass of Water First, we need to calculate the mass of water. We know that ma ss = d e n s i t y × v o l u m e Since the density of water is 1.0 g / m L and the volume is 450 m L , the mass of water is: ma s s w a t er = 1.0 m L g × 450 m L = 450 g We will also need the mass of water in kg later, so let's convert it now: ma s s w a t er = 450 g × 1000 g 1 k g = 0.45 k g
Calculating Freezing Point Depression Next, we calculate the freezing point depression, Δ T f . The freezing point of pure water is 0 ∘ C , and the desired freezing point of the solution is − 15. 0 ∘ C . Therefore, Δ T f = T f , p u re − T f , so l u t i o n = 0 ∘ C − ( − 15. 0 ∘ C ) = 15. 0 ∘ C
Calculating the Molality of the Solution Now we use the freezing point depression formula to find the molality of the solution: Δ T f = K f × m where m is the molality. We can solve for m :
m = K f Δ T f = 1.8 6 ∘ C / m 15. 0 ∘ C ≈ 8.0645 m
Calculating Moles of Ethanol Molality is defined as moles of solute (ethanol) per kilogram of solvent (water): m = ma s s w a t er ( ink g ) m o l e s e t han o l We can solve for the moles of ethanol: m o l e s e t han o l = m × ma s s w a t er ( ink g ) = 8.0645 k g m o l × 0.45 k g ≈ 3.629 m o l es
Calculating Mass of Ethanol Now we need to find the mass of ethanol using its molar mass. The molar mass of ethanol ( C 2 H 5 O H ) is approximately 2 ( 12.01 ) + 6 ( 1.008 ) + 16.00 = 46.068 g / m o l . Therefore, ma s s e t han o l = m o l e s e t han o l × m o l a r ma s s e t han o l = 3.629 m o l × 46.068 m o l g ≈ 167.2 g
Calculating Volume of Ethanol Finally, we can calculate the volume of ethanol using its density: v o l u m e e t han o l = d e n s i t y e t han o l ma s s e t han o l = 0.7893 c m 3 g 167.2 g ≈ 211.8 c m 3 Since 1 c m 3 = 1 m L , the volume of ethanol is approximately 211.8 m L .
Final Answer Therefore, the volume of ethanol that should be added to 450 mL of water to have a solution that freezes at − 15. 0 ∘ C is approximately 212 mL.
Examples
Freezing point depression is a colligative property, meaning it depends on the concentration of solute particles rather than the identity of the solute. This principle is used in real life to de-ice roads during winter. Salt (typically NaCl or C a C l 2 ) is spread on the roads, which dissolves in the thin layer of water present and lowers the freezing point, preventing ice formation. The amount of salt needed depends on the expected temperature drop and the amount of water present. Similarly, antifreeze (ethylene glycol) is added to car radiators to prevent the water in the cooling system from freezing in cold weather, which could damage the engine.
