Question

Question #38e70

Answer 1

Here's what I got.

Explanation:

Since all you have to do here is use the equation for boiling-point elevation, I'll show you how to solve part (A), and you can solve the other parts as practice.

Now, you can't really provide a clear answer using the information given. Here's why that is so.

The equation for boiling-point elevation looks like this

`color(blue)(DeltaT_b = i * K_b * b)" "`, where

`DeltaT_b` - the boiling-point elevation;
`i` - the van't Hoff factor
`K_b` - the ebullioscopic constant of the solvent;
`b` - the molality of the solution.

The ebullioscopic constant for water is equal to `0.512^@"C kg mol"^(-1)`

http://www.vaxasoftware.com/doc_eduen/qui/tcriosebu.pdf

As you can see, an important piece of information is missing. More specifically, you don't know if the solute is an electrolyte or a non-electrolyte.

The van't Hoff factor, `i`, tells you the ratio between the number of moles of solute you dissolve in solution and the number of moles of particles produced when the solute dissolves.

For example, glucose, `"C"_6"H"_12"O"_6`, is a non-electrolyte, which means that it dissolves in aqueous solution without breaking up into ions. For non-electrolytes, the van't Hoff factor is equal to `1`.

On the other hand, sodium chloride, `"NaCl"`, is an electrolyte, which means that it dissociates completely in aqueous solution to form sodium cations and chloride anions

`"NaCl"_text((aq]) -> "Na"_text((aq])^(+) + "Cl"_text((aq])^(-)`

For sodium chloride, the van't Hoff factor will be equal to `2` because one mole of sodium chloride produces two moles of ions in solution.

I'll do the calculations using both values of the van't Hoff factor.

For the glucose solution, the boiling-point elevation will be

`DeltaT_b = 1 * 0.512 ^@"C" color(red)(cancel(color(black)("kg"))) color(red)(cancel(color(black)("mol"^(-1)))) * 0.325 color(red)(cancel(color(black)("mol"))) color(red)(cancel(color(black)("kg"^(-1))))`

`DeltaT_b = 0.1664^@"C"`

Boiling-point elevation is defined as

`color(blue)(DeltaT_b = T_b - T_b^@)" "`, where

`T_b` - the boiling point of the solution
`T_b^@` - the boiling point of the pure solvent

In your case, you will get

`T_b = DeltaT_b + T_b^@ = 0.1664^@"C" + 100.0^@"C" = color(green)(100.166^@"C")`

FOr the sodium chloride solution, the boiling-point elevation will be

`DeltaT_b = 2 * 0.512 ^@"C" color(red)(cancel(color(black)("kg"))) color(red)(cancel(color(black)("mol"^(-1)))) * 0.325 color(red)(cancel(color(black)("mol"))) color(red)(cancel(color(black)("kg"^(-1))))`

`DeltaT_b = 0.3328^@"C"`

The boiling point of the solution will be

`T_b = 0.3328^@"C" + 100.0^@"C" = color(green)(100.333^@"C")`

The answers are rounded to three decimal places.