Question #0c4c5
1 Answer
Explanation:
When dealing with the freezing point of a solution, your first goal is to determine its molality by
- determining the number of moles of solute present in the solution
- determining the mass of the solvent in kilograms
Once you know the molality of the solution, you can find the freezing-point depression by using the equation
#color(blue)(|bar(ul(color(white)(a/a)DeltaT_f = i * K_f * bcolor(white)(a/a)|)))" "# , where
In your case, the cryoscopic constant of water is said to be
#K_f = 1.86^@"C kg mol"^(-1)#
Glucose is a non-electrolyte, i.e. it remains undissociated in aqueous solution, which means that its van't Hoff factor will be
So, you know that your solution contains
#53.4 color(red)(cancel(color(black)("g"))) * "1 mole glucose"/(180.16color(red)(cancel(color(black)("g")))) = "0.2964 moles glucose"#
The solution contains
#475color(red)(cancel(color(black)("g"))) * "1 kg"/(10^3color(red)(cancel(color(black)("g")))) = "0.475 kg"#
The molaity of the solution, which is defined as
#color(blue)(|bar(ul(color(white)(a/a)"molality" = "moles of solute"/"kilogram of solvent"color(white)(a/a)|)))#
will thus be
#b = "0.2964 moles"/"0.475 kg" = "0.624 mol kg"^(-1)#
Now plug in your values and solve for the freezing-point depression
#DeltaT_f = 1 * 1.86^@"C" color(red)(cancel(color(black)("kg")))color(red)(cancel(color(black)("mol"^(-1)))) * 0.624 color(red)(cancel(color(black)("mol"))) color(red)(cancel(color(black)("kg"^(-1))))#
#DeltaT_f = 1.161^@"C"#
The freezing-point depression is defined as
#color(blue)(|bar(ul(color(white)(a/a)DeltaT_f = T_f^@ - T_"f sol"color(white)(a/a)|)))" "# , where
This means that the freezing point of the solution will be
#T_"f sol" = T_f^@ - DeltaT_f#
#T_"f sol" = 0^@"C" - 1.161^@"C" = color(green)(|bar(ul(color(white)(a/a)color(black)(-1.16^@"C")color(white)(a/a)|)))#
The answer is rounded to three sig figs.