Question #cb01d
1 Answer
Explanation:
Your tool of choice here will be the equation that helps you find the boiling-point elevation of the solution
#color(blue)(|bar(ul(color(white)(a/a)DeltaT_b = i * K_b * bcolor(white)(a/a)|)))#
Here
You already know the molality of the solution, which is given as
#"0.38 m" = "0.38 mol kg"^(-1)#
and the ebullioscopic constant of water, which is given as
You can rewrite the ebullioscopic constant of water in terms of degrees Celsius by using the fact that an increase of
You will thus have
#K_b = "0.51"""^@"C kg mol"^(-1)#
Now, focus on find the van't Hoff factor. Calcium chloride,
#"CaCl"_ (color(red)(2)(aq)) -> "Ca"_ ((aq))^(2+) + color(red)(2)"Cl"_ ((aq))^(-)#
Notice that every mole of calcium chloride that dissolves in solution produces
The van't Hoff factor tells you the ratio that exists between the number of moles of solute dissolved and the number of moles of particles of solute produced in solution.
In this case, one mole of solute produces
#"1 mole Ca"^(2+) + color(red)(2)color(white)(a)"moles Cl"^(-) = "3 moles of ions"#
Therefore, the van't Hoff factor will be equal to
Plug in your values into the equation and solve for the boiling-point elevation of the solution
#DeltaT_b = 3 * 0.51""^@"C" color(red)(cancel(color(black)("kg"))) color(red)(cancel(color(black)("mol"^(-1)))) * 0.38 color(red)(cancel(color(black)("mol"))) color(red)(cancel(color(black)("kg"^(-1))))#
#DeltaT_b = color(green)(|bar(ul(color(white)(a/a)color(black)(0.58^@"C")color(white)(a/a)|)))#
The answer is rounded to two sig figs.
Pure water boils at a temperature of
#T_b = 100^@"C" + 0.58^@"C" = 100.58^@"C"#