A solution that contains 55.0 g of ascorbic acid (Vitamin C) in 250. g of water freezes at -2.34^$\circ$C. Calculate the molar mass (in units of g/mol) of the solute. K_f of water is 1.86^$\circ$C/m. A) 1.26 B) 10.9 C) 43.6 D) 175 E) 277

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The freezing point depression formula is ΔTf = i * Kf * m, where ΔTf is the change in freezing point, i is the van't Hoff factor (which is 1 for non-electrolytes like ascorbic acid), Kf is the cryoscopic constant of the solvent (water in this case, 1.86 °C/m), and m is the molality of the solution. Given that the solution freezes at -2.34 °C, and the freezing point of pure water is 0 °C, ΔTf = 2.34 °C. To find the molality, we rearrange the formula to m = ΔTf / (i*Kf) = 2.34 °C / (1*1.86 °C/m) = 1.258 mol/kg. Since there are 55.0 g of ascorbic acid in 250 g (or 0.250 kg) of water, the number of moles of ascorbic acid is molality (m) * kg of solvent = 1.258 mol/kg * 0.250 kg = 0.3145 mol. The molar mass of ascorbic acid is then the mass of ascorbic acid divided by the number of moles: 55.0 g / 0.3145 mol = 174.9 g/mol, which rounds to 175 g/mol.

A Solution That Contains 55 ∘\circ∘ C Calculate the Molar Mass (In Units of G/mol) of C)

A Solution That Contains 55 $\circ$ C Calculate the Molar Mass (In Units of G/mol) of C)