Benzaldehyde (? = 106.1 g/mol), also known as oil of almonds, is used in the manufacture of dyes and perfumes and in flavorings. What would be the freezing point of a solution prepared by dissolving 75.00 g of benzaldehyde in 850.0 g of ethanol? K_f = 1.99$\degree$C/m, freezing point of pure ethanol = -117.3$\degree$C.

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First, we need to calculate the molality of the solution using the formula: molality = moles of solute / mass of solvent (in kg) To find the moles of benzaldehyde, we divide the given mass by the molar mass: moles of benzaldehyde = 75.00 g / 106.1 g/mol = 0.707 mol Then, we convert the mass of ethanol to kg: mass of ethanol = 850.0 g = 0.850 kg Now, we can calculate the molality: molality = 0.707 mol / 0.850 kg = 0.831 mol/kg Next, we need to calculate the freezing point depression using the formula: ΔTf = Kf x molality where Kf is the freezing point depression constant for ethanol (given as K_f = 1.99F1F1F1S1F1F1F10C/m). ΔTf = 1.99F1F1F1S1F1F1F10C/m x 0.831 mol/kg = 1.656F1F1F1S1F1F1F10C Finally, we can calculate the freezing point of the solution by subtracting the ΔTf from the freezing point of pure ethanol: freezing point of solution = -117.3F1F1F1S1F1F1F10C - 1.656F1F1F1S1F1F1F10C = -119.0F1F1F1S1F1F1F10C Therefore, the best choice is C, -119.0F1F1F1S1F1F1F10C.

Benzaldehyde (? = 106 °\degree° C/m, Freezing Point of Pure Ethanol = -117

Benzaldehyde (? = 106 $\degree$ C/m, Freezing Point of Pure Ethanol = -117