A solution of a nonelectrolyte solution contains 30.0 g of solute dissolved in 250.0g of water.The freezing point of the water is observed to be -2.50˚C.The K_f for water is 1.86 ˚C/m and normal freezing point of water is 0.00˚C.What is the molar mass of the substance? A)335 g/mol B)89.5 g/mol C)895 g/mol D)33.5 g/mol

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The freezing point depression ($\Delta T_f$) is given by the formula $\Delta T_f = i \cdot K_f \cdot m$, where $i$ is the van't Hoff factor (which is 1 for a nonelectrolyte), $K_f$ is the cryoscopic constant for the solvent (1.86 ˚C/m for water), and $m$ is the molality of the solution. Given that the observed freezing point of the solution is -2.50˚C and the normal freezing point of water is 0.00˚C, the freezing point depression is 2.50˚C.First, calculate the molality ($m$) of the solution:$$2.50 = 1 \cdot 1.86 \cdot m$$$$m = \frac{2.50}{1.86} = 1.344 \, \text{mol/kg}$$Since there are 30.0 g of solute in 250.0 g (or 0.250 kg) of water, we can find the moles of solute:$$1.344 \, \text{mol/kg} = \frac{\text{moles of solute}}{0.250 \, \text{kg}}$$$$\text{moles of solute} = 1.344 \times 0.250 = 0.336 \, \text{mol}$$Finally, calculate the molar mass of the solute:$$Molar \, mass = \frac{mass \, of \, solute}{moles \, of \, solute} = \frac{30.0 \, g}{0.336 \, mol} = 89.3 \, g/mol$$However, given the choices provided and rounding, the closest answer to the calculated molar mass is 89.5 g/mol, which seems to be a mistake in the calculation or interpretation of the choices. The correct process leads to an answer close to 89.5 g/mol, but none of the choices exactly match the calculated value. Given the context, it appears there might have been a misunderstanding in the calculation or the presentation of the answer choices. The correct calculation process was intended to lead to the identification of the molar mass based on the freezing point depression, but the final step or choice selection seems to have been misaligned with the calculated value.

A Solution of a Nonelectrolyte Solution Contains 30