Question #9d5e0
1 Answer
Explanation:
The first thing to do here is calculate the freezing-point depression,
Once you know that, you can use the freezing point of the pure solvent to find the freezing point of the solution.
So, the freezing-point depression is calculated using the equation
#color(blue)(|bar(ul(color(white)(a/a)DeltaT_f = i * K_f * bcolor(white)(a/a)|)))" "# , where
The cryoscopic constant of water is equal to
#K_f = 1.86^@"C kg mol"^(-1)#
Now, sodium chloride,
#"NaCl"_ ((aq)) -> "Na"_ ((aq))^(+) + "Cl"_ ((aq))^(-)#
Notice that every mole of sodium chloride added to the solution produces
This means that the van't Hoff factor for sodium chloride, which essentially tells you how many moles of particles of solute are produced in solution per mole of solute dissolved, will be equal to
Plug in your values to find the freezing-point depression of the solution
#DeltaT_f = 2 * 1.86""^@"C" color(red)(cancel(color(black)("kg"))) * color(red)(cancel(color(black)("mol"^(-1)))) * 0.50color(red)(cancel(color(black)("mol"))) color(red)(cancel(color(black)("kg"^(-1))))#
#DeltaT_f = 1.86^@"C"#
The freezing-point depression is defined as
#color(purple)(|bar(ul(color(white)(a/a)color(black)(DeltaT_f = T_f^@ - T_"f sol")color(white)(a/a)|)))" "#
Here
Pure water freezes at
#DeltaT_"f sol" = 0^@"C" - 1.86^@"C" = color(green)(|bar(ul(color(white)(a/a)color(black)(-1.9^@"C")color(white)(a/a)|)))#
The answer is rounded to two sig figs, the number of sig figs you have for the molality of the solution.