# Question #ec1e7

##### 1 Answer

#### Explanation:

The idea here is that you need to use the difference between the freezing point of this solution and the freezing point of *pure cyclohexane* to determine the molality of the solution.

This, in turn, will help you find the number of moles of solute.

So, you know that the equation that links *freezing-point depression* and *molality* looks like this

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

*van't Hoff factor*, equal to

**cryoscopic constant** of the solvent;

The freezing point of *pure cyclohexane* is

Rearrange the equation and solve for

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

In this case, the freezing-point depression will be

#DeltaT_f = T_f^@ - T_"f sol"#

#DeltaT_f = 6.47""^@"C" - 2.79""^@"C" = 3.98""^@"C"#

This means that

#b = (3.98color(red)(cancel(color(black)(""^@"C"))))/(1 * "20.2 kg"color(red)(cancel(color(black)(""^@"C")))"mol"""^(-1)) = "0.1970 mol/kg"#

Since molality is defined as moles of solute per **kilograms** of solvent, you can say that

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

This means that you have

#n_"solute" = 0.1970"mol"/color(red)(cancel(color(black)("kg"))) * 20.0 * 10^(-3)color(red)(cancel(color(black)("kg"))) = "0.00394 moles"#

Use the compound's known mass to find the value of its molar mass

#M_M = m/n = "0.627 g"/"0.00394 moles" = color(green)("159 g/mol")#

The answer is rounded to three sig figs.