1.80 g of solute A was dissolved in 62.5 cm$$^3$$ of ethanol and freezing point of the solution was found to be 155.1 K. The molar mass of solute A is _____ g mol$$^{-1}$$. [Given: Freezing point of ethanol is 156.0 K. Density of ethanol is 0.80 g cm$$^{-3}$$. Freezing point depression constant of ethanol is 2.00 K kg mol$$^{-1}$$]

We have 1.80 g of solute A dissolved in 62.5 cm$$^3$$ of ethanol. The freezing point of the solution is 155.1 K, while the freezing point of pure ethanol is 156.0 K. We need to find the molar mass of solute A.

We begin by calculating the freezing point depression:

$$\Delta T_f = T_f^0 - T_f = 156.0 - 155.1 = 0.9 \text{ K}$$

Now we find the mass of ethanol (solvent). Given the volume is 62.5 cm$$^3$$ and the density is 0.80 g cm$$^{-3}$$:

$$\text{Mass of ethanol} = 62.5 \times 0.80 = 50.0 \text{ g} = 0.050 \text{ kg}$$

The freezing point depression is related to molality by:

$$\Delta T_f = K_f \times m$$

where $$m$$ is the molality (moles of solute per kg of solvent) and $$K_f = 2.00$$ K kg mol$$^{-1}$$. Substituting:

$$0.9 = 2.00 \times m$$

$$m = \frac{0.9}{2.00} = 0.45 \text{ mol kg}^{-1}$$

The molality is defined as moles of solute per kg of solvent, so the number of moles of solute A is:

$$n_A = m \times \text{mass of solvent (in kg)} = 0.45 \times 0.050 = 0.0225 \text{ mol}$$

Now the molar mass of solute A is:

$$M_A = \frac{\text{mass of solute}}{n_A} = \frac{1.80}{0.0225} = 80 \text{ g mol}^{-1}$$

Hence, the correct answer is 80.