# Methanol, #CH_3OH# is used as an antifreeze. How many grams of methanol would you need per 1000.0 g of water for an aqueous solution to stay liquid at -20.0° C?

##### 1 Answer

#### Answer:

#### Explanation:

Your strategy here will be to use the equation for *freezing-point depression* to determine the molality of the target solution, then use methanol's *molar mass* to find the mass of the compound needed to make this solution.

The equation that describes *freezing-point depression* looks like this

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

*van't Hoff factor*

*cryoscopic constant* of the solvent;

The cryoscopic constant of water is equal to

http://www.vaxasoftware.com/doc_eduen/qui/tcriosebu.pdf

Now, since you're dealing with a **non-electrolyte**, the *van't Hoff factor*, which tells you what the ratio between the number of moles of solute dissolved in water and the number of moles of particles produced, will be equal to

The freezing-point depression is defined as

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

*pure solvent*

*Pure water* freezes at

#DeltaT_f = 0^@"C" - (-20.0^@"C") = 20.0^@"C"#

Rearrange the equation for freezing-point depression and solve for

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

#b = (20.0 color(red)(cancel(color(black)(""^@"C"))))/(1 * 1.86 color(red)(cancel(color(black)(""^@"C"))) "kg mol"^(-1)) = "10.753 mol kg"^(-1)#

As you know, molality is defined as moles of solute per **kilograms of solvent**. Use the given mass of water to find how many moles of methanol must be present in the solution

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

#n = "10.753 mol" color(red)(cancel(color(black)("kg"^(-1)))) * 1000.0 * 10^(-3)color(red)(cancel(color(black)("kg")))#

#n = "10.753 moles CH"_3"OH"#

Finally, use methanol's molar mass to find how many grams would contain that many moles

#10.753 color(red)(cancel(color(black)("moles CH"_3"OH"))) * "32.042 g"/(1color(red)(cancel(color(black)("mole CH"_3"OH")))) = "344.55 g"#

Rounded to three sig figs, the number of sig figs you have for the freezing temperature of the solution

#m_(CH_3OH) = "355 g"#

So, a solution that contains **bigger** than this value

#m_(CH_3OH) > color(green)("355 g")#