# 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

*van't Hoff factor*

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 **equivalent** to 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, **soluble** ionic compound that dissociates completely in aqueous solution to produce calcium cations,

#"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 **mole** of calcium cations and **moles** of chloride anions.

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"#