After you preform your experiment, you determine that the Kf value for naphthalene is 6.91 °C/m . You are using 10g of naphthalene and added 1.0 g of your unknown. The the freezing point of the solvent decreased by 4.49 °C°C when the unknown was added. Knowing this information, determine the molar mass of the unknown.

153.9 g/mol

Explanation:

The freezing point depression for an ideal solution is given by the expresion:

ΔTf = Kf m

where:

ΔTf is the freezing point depression

Kf is the freezing constant of the solvent

m is the molalility  which mol of solute per kilogram solvent

The moles of solute is the mass divided into its molecular weight, MW

substituting into the equation, we have:

4.49 º C =  6.91 ºC/m x [(1.0 g/MW) / 0.010 kg]

4.49 º C =  6.91 ºC/mol/kg x  [(1.0 g/MW) / 0.010 kg]

⇒4.49 = 691 g/mol/MW

⇒ MW = 691 / 4.49 g/mol = 153.9 g /mol

the working equation for the angular size is the product of the physical size and 360 degrees divided by the product of two pi and distance from earth. the minimum size would be 0.559 degrees while the maximum size would be 0.49 degrees.