# Question #da7b4

##### 1 Answer

#### Answer:

#### Explanation:

Your tool of choice here will be the equation

#color(blue)(|bar(ul(color(white)(a/a)DeltaT_f = i * K_f * bcolor(white)(a/a)|)))#

Here

*van't Hoff factor*

*cryoscopic constant* of the solvent;

This equation will allow you to find the **freezing-point depression**, *freezing point* of the solution.

Your solute, naphthalene, **non-electrolyte**, which means that the *van't Hoff factor* for this solution will be

The first thing to do here is calculate the **molality** of the solution by figuring out

how manymoles of soluteyou have presenthow manykilograms of solventyou have in your solution

The problem provides you the **molar masses** of the elements that make up naphthalene, so calculate the molar mass of the compound by

#M_("M C"_10"H"_8) = 10 xx "12.011 g mol"^(-1) + 8 xx "1.008 g mol"^(-1)#

#M_("M C"_8"H"_10) = "128.174 g mol"^(-1)#

The number of moles of naphthalene present in the sample is equal to

#0.134 color(red)(cancel(color(black)("g"))) * ("1 mole C"_10"H"_8)/(128.174 color(red)(cancel(color(black)("g")))) = "0.001045 moles C"_10"H"_8#

The mass of cyclohexane, *kilograms*, is equal to

#5.00 color(red)(cancel(color(black)("g"))) * "1 kg"/(10^3color(red)(cancel(color(black)("g")))) = 5.00 * 10^(-3)"kg"#

The molality of the solution will be

#b = "0.001045 moles"/(5.00 * 10^(-3)"kg") = "0.209 mol kg"^(-1)#

Next, use your values to calculate

#DeltaT_f = 1 * 20.0^@"C" color(red)(cancel(color(black)("kg"))) color(red)(cancel(color(black)("mol"^(-1)))) * 0.209 color(red)(cancel(color(black)("mol"))) color(red)(cancel(color(black)("kg"^(-1))))#

#DeltaT_f = 4.18^@"C"#

Now, a solution's **freezing-point depression** essentially tells you how the freezing point of the solution, *pure solvent*,

More specifically, the freezing-point depression tells you how many degrees **lower** the freezing point of the solution will be compared to that of the pure solvent

#color(purple)(|bar(ul(color(white)(a/a)color(black)(T_"sol" = T_f^@ - DeltaT_f)color(white)(a/a)|)))#

In your case, you have

#T_"sol" = 6.60^@"C" - 4.18^@"C" = color(green)(|bar(ul(color(white)(a/a)color(black)(2.42^@"C")color(white)(a/a)|)))#

The answer is rounded to two *decimal places*.