Question #da7b4
1 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
This equation will allow you to find the freezing-point depression,
Your solute, naphthalene,
The first thing to do here is calculate the molality of the solution by figuring out
- how many moles of solute you have present
- how many kilograms of solvent you 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,
#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,
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.