# A solution containing 10.0 g of an unknown liquid and 90.0 g water has a freezing point of -3.33° C. Given #K_f# = 1.86°C/m for water, what is the molar mass of the unknown liquid?

##### 1 Answer

#### Explanation:

The most important thing to realize here is that since the solute is a **liquid**, you must be dealing with a **covalent compound**, which as you know **does not** ionize in aqueous solution.

This implies that the *van't Hoff factor*, which tells you the ratio between the **number of moles of particles** produced in solution and the number of moles of solute dissolved, will be equal to

Simply put, *every mole* of your unknown compound will produce

Your tool of choice here will be the equation for *freezing-point depression*, which looks like this

#color(blue)(DeltaT_f = i * K_f * b)" "# , where

*van't Hoff factor*

*cryoscopic constant* of the solvent;

The freezing-point depression is defined as

#color(blue)(DeltaT_f = T_f^@ - T_f)" "# , where

**pure solvent**

Use the given freezing point of the solution to find

#DeltaT_f = 0^@"C" - (-3.33^@"C") = 3.33^@"C"#

Now, you strategy here will be to find the molaity of the solution, *number of moles* of solute present in this solution.

So, rearrange the above equation to solve for *cryoscopic constant* for water,

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

#b = (3.33 color(red)(cancel(color(black)(""^@"C"))))/(1 * 1.86 color(red)(cancel(color(black)(""^@"C"))) "kg mol"^(-1)) = "1.7903 mol kg"^(-1)#

Now, a solution's molality is defined as the number of moles of solute divided by the mass of the solvent - expressed in **kilograms**.

#color(blue)(b = n_"solute"/m_"solvent in kg")#

Since your solution is said to contain

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

You will have

#n_"solute" = "1.7903 mol" color(red)(cancel(color(black)("kg"^(-1)))) * 90.0 * 10^(-3)color(red)(cancel(color(black)("kg"))) = "0.16113 moles"#

Finally, the **molar mass** of a substance can be found by dividing the mass of a given sample of that substance by the total number of moles it contains.

#color(blue)(M_M = m/n)#

In your case, the molar mass of the liquid will be

#M_M = "10.0 g"/"0.16113 moles" = "62.062 g mol"^(-1)#

Rounded to three sig figs, the answer will be

#M_M = color(green)("62.1 g mol"^(-1))#