# Question #5e092

##### 1 Answer

The molar mass of the compound is

#### Explanation:

The way I see it, the only reasonable thing to determine here is the *molar mass* of the unknown compound.

The idea is that you can use the freezing point of the solution to find its molality. The equation for *freezing-point depression* looks like this

#DeltaT_f = i * K_f * b" "# , where

*van't Hoff factor*, equal to

*cryoscopic constant* of the solvent;

The cryoscopic constant of water is equal to

#DeltaT_f = T_f^@ - T_f" "# , where

*pure solvent*.

In your case, you have

#DeltaT_f = 0""^@"C" - (-1.3""^@"C") = 1.3""^@"C"#

This means that the molality of the solution will be

#b = (DeltaT_f)/(i * K_f)#

#b = (1.3color(red)(cancel(color(black)(""^@"C"))))/(1 * 1.853color(red)(cancel(color(black)(""^@"C"))) "Kg mol"""^(-1)) = "0.7016 molal"#

Molality is defined as moles of solute, in your case the unknown non-electrolyte, divided by the mass of the solvent - expressed in **kilograms**.

Use the solution's molality to find the number of moles of solute

#b = n_"solute"/m_"solvent" implies n_"solute" = b * m_"solvent"#

#n_"solute" = 0.7016"mol"/color(red)(cancel(color(black)("kg"))) * 156.5 * 10^(-3)color(red)(cancel(color(black)("kg"))) = "0.1098 moles"#

This means that the molar mass of the compound will be

#M_M = m/n = "35.9 g"/"0.1098 moles" = "326.9 g/mol"#

Rounded to two sig figs, the answer will be

#M_M = color(green)("330 g/mol")#

**SIDE NOTE** *If this is not what you are looking for, let me know, I will wdit the answer accordingly*.