Question

Question #59807

Answer 1

`114.97^"o""C"`

Explanation:

We're asked to find the new freezing point of the ethanol if `0.50 "g NaCl"` is dissolved in `1.0"kg"` of ethanol.

This problem deals with colligative properties of solutions, this one in particular is freezing-point depression. The formula to use for freezing-point depression is

`DeltaT_"f" = i "•" m "•" K_"f"`

where

• `DeltaT_"f"` is the change in freezing point temperature of the solution

• `i` is called the van't Hoff factor, which for these purposes is just the number of dissolved ions in solution (`2` in this case: `"Na"^+` and `"Cl"^-`)

• `m` is the molality of the solution, which is the moles of solute per kilogram of solvent. To find this, we need to convert the given mass of `"NaCl"` to moles, using its molar mass:

`"mol NaCl" = 0.50cancel("g NaCl")((1"mol NaCl")/(58.44cancel("g NaCl")))`

`= color(red)(0.00856` `color(red)("mol NaCl"`

The molality is thus

`color(red)(0.00856"mol NaCl")/(1"kg ethanol") = color(green)(0.00856 "mol"/"kg"`

• `K_"f"` is the molal freezing point constant for the solvent (ethanol), which is given: `1.99 (""^"o""C")/(m)`

Plugging in out known quantities, we have

`DeltaT_"f" = (2)(0.00856cancel(m))(1.99 (""^"o""C")/(cancel(m)))`

`= 0.034^"o""C"`

This is how much the freezing point decreases. We're asked to find its actual freezing point. Doing this is simple enough, we just subtract this value from ethanol's regular freezing point (`115^"o""C"`):

`115^"o""C" - 0.034^"o""C" = color(blue)(114.97^"o""C"`

The new freezing point is thus `114.97 ^"o""C"`.

(If you follow the technical rules of significant figures, the result is still `115^"o""C"`, as if nothing changed. This is because the value `115^"o""C"` (its regular freezing point) has no decimal places, and subtracting `0.034^"o""C"` from it would still yield `115^"o""C"` (rounded to `0` decimal places). The concentration of `"NaCl"` in solution is so small that it practically has no effect on the freezing point.)