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A solution was made by adding 800 g of ethanol, C₂H₅OH, to 8.0x10³g of water. How much would this lower the freezing point? The freezing point depression constant, kf, of water is 1.86 °C/m and the molar mass of ethanol is 46.08 g/mol
A solution was made by adding 800 g of ethanol, C₂H₅OH, to 8.0x10³g of water. How much would this lower the freezing point? The freezing point depression constant, kf, of water is 1.86 °C/m and the molar mass of ethanol is 46.08 g/mola) 3.2°C
b) 4.1°C
c) 8.2°C
d) 16°C
Final answer:The freezing point of the solution is lowered by approximately 4.04°C when 800 g of ethanol is added to 8.0x10³ g of water, correlating closest to option b) 4.1°C. Therefore, the correct option is b)Explanation:The question concerns the calculation of freezing point depression when ethanol is added to water. To determine how much the freezing point is lowered, we can use the following formula:ΔTf = Kf × m × i, where ΔTf is the change in freezing point, Kf is the freezing point depression constant of the solvent (water in this case), m is the molality of the solution, and i is the van 't Hoff factor (which is 1 for non-electrolytes like ethanol).First, we calculate the molality (m) of the solution by dividing the moles of ethanol by the kilograms of water. The moles of ethanol are found by dividing its mass by its molar mass:Moles of ethanol = 800 g / 46.08 g/mol = 17.36 molMolality (m) = 17.36 mol / 8.0 kg = 2.17 mNext, we apply the formula:ΔTf = 1.86 °C/kg·mol × 2.17 m × 1 = 4.04 °CTherefore, the freezing point is lowered by approximately 4.04 °C, which is closest to option b) 4.1°C....