Question #4b1a8
1 Answer
Explanation:
Your tool of choice here will be the equation
#color(blue)(ul(color(black)(DeltaT = i * K_f * b)))#
Here
Now, water has a cryoscopic constant equal to
#K_f = 1.86^@"C kg mol"^(-1)#
http://www.vaxasoftware.com/doc_eduen/qui/tcriosebu.pdf
Calcium chloride,
#"CaCl"_ (2(aq)) -> "Ca"_ ((aq))^(2+) + 2"Cl"_ ((aq))^(-)#
Notice that every mole of calcium chloride that dissolves in water produces
- one mole of calcium cations,
#1 xx "Ca"^(2+)# - two moles of chloride anions,
#2 xx "Cl"^(-)#
The van't Hoff factor tells you the ratio that exists between the number of moles of solute dissolved in solution and the number of moles of particles of solute produced in solution.
In this case,
#i = 3 -># one mole of calcium chloride dissolved, three moles of ions produced
Now, water has a normal freezing point of
#color(blue)(ul(color(black)(DeltaT = T_"f pure water" - T_"f solution")))#
#DeltaT = 0^@"C" - (-4.2^@"C")#
#DeltaT = 4.2^@"C"#
Rearrange the above equation and solve for
#DeltaT = i * K_f * b implies b = (DeltaT)/(i * K_f)#
Plug in your values to find
#b = (4.2 color(red)(cancel(color(black)(""^@"C"))))/(3 * 1.86 color(red)(cancel(color(black)(""^@"C"))) "kg mol"^(-1)) = "0.753 mol kg"^(-1)#
The molality of the solution is calculated by taking the number of moles of solute present in
In this case, your solution contains
#750. color(red)(cancel(color(black)("g"))) * "1 kg"/(10^3color(red)(cancel(color(black)("g")))) = "0.750 kg"#
of water, which means that it also contains
#0.750 color(red)(cancel(color(black)("kg water"))) * "0.753 moles CaCl"_2/(1color(red)(cancel(color(black)("kg water")))) = "0.565 moles CaCl"_2#
To convert this to grams, use the molar mass of calcium chloride
#0.565 color(red)(cancel(color(black)("moles CaCl"_2))) * "110.98 g"/(1color(red)(cancel(color(black)("mole CaCl"_2)))) = color(darkgreen)(ul(color(black)("63 g")))#
The answer is rounded to two sig figs, the number of sig figs you have for the freezing point of the solution.