A solution that contains 55.0 g of ascorbic acid (vitamin C)in 250.g of water freezes at -2.34°C.Calculate the molar mass (in units of g/mol)of the solute. (For water,K_f = 1.86°C/m) A)1.26 g/mol B)10.9 g/mol C)43.6 g/mol D)175 g/mol E)277 g/mol

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The freezing point depression formula is ΔTf = i * Kf * m. Here, ΔTf = 2.34°C, Kf = 1.86°C/m, and i = 1 (since ascorbic acid is a non-electrolyte). Solving for molality (m), we get m = 2.34 / 1.86 = 1.258 m. The molality is moles of solute per kg of solvent, so moles of solute = 1.258 m * 0.250 kg = 0.3145 moles. The molar mass is mass of solute / moles of solute = 55.0 g / 0.3145 moles = 175 g/mol.

A Solution That Contains 55