Question

How many grams of S were used? What is the freezing point of S in degrees C?

A solvent, S, having a density of 1.06 g/mL and Kf = 3.90 ºC/m was uniformly cooled. A graph of temperature-time readings showed a plateau (flat) region [S(l)/S(s)] at 16.7 ºC. A solution of 2.49 g compound B was dissolved in 21.6 mL of S. I was cooled and dropped in temperature until reaching 3.38 degrees C.

Answer 1

`"Mass of Solvent"` = `"22.896 g"`

`"Freezing point = 16.7 ºC"`

Explanation:

`"Mass of Solvent"` = `"Volume x density"`

`"Mass of Solvent"` = `"21.6 mL" (1.06 g/(mL))` = 22.896 g
Or,

`"Mass of Solvent"` = `"0.023 kg"`

The Freezing point is read from the plateau zone which represents the Solid-Liquid Solvent equilibrium:

`"Freezing point = 16.7 ºC"`

IN ADDITION `=>` BLAGDEN'S LAW
evaluates the Freezing Point Solvent Depression (`Delta T`) as function of its Solute Concentration (for an Ideal Solution):

`"Delta T` = `i K M`

where,

`i` : Van't Hoff Factor

`K` : Cryoscopic constant `(("ºC kg")/(mol))`
`M` : Molality of solute `("mol of solute"/("kg of solvent"))`

Assuming that is not an electrolyte; i.e. "i = 1"

`(16.7 - 3.38) ºC` = `(1) (3.9 ºC)` `(("mol of solute")/("0.022896 kg"))`

`"mol of solute"` = 0.0782

`"Molecular Weight of solute"` = 2.29 g / 0.0782 mol = 29.28 g/mol