Question

A solution of water (kf=1.86 ∘c/m) and glucose freezes at − 2.75 ∘c. what is the molal concentration of glucose in this solution? assume that the freezing point of pure water is 0.00 ∘c.

Answer 1

Step-by-step explanation:

Relation between freezing temperature and molal concentration is as follows.

`\Delta T_(f) = k_(f) * m`

The given data is as follows.

`\Delta T_(f)` = difference in temperature =
`[0 - (-2.75)]^(o)C` =
`2.75^(o)C`

`k_(f) = 1.86^(o)C/mol`

molality, (m) = ?

Now, putting the given values into the above formula as follows.

m =
`(\Delta T_(f))/(k_(f))`

=
`(2.75^(o)C)/(1.86^(o)C/mol)`

= 1.48 m

Therefore, we can conclude that molal concentration of glucose in the given solution is 1.48 m.

Answer 2

Answer is: the molal concentration of glucose in this solution is 1,478 m.
Tf(glucose) = -2,75°C.
Tf(water) = 0°C.
ΔT(solution) = 2,75°C.
Kf(water) = 1,86°C/m.
ΔT = Kf(water) · b(solution).
b(solution) = ΔT ÷ Kf(water).
b(solution) = 2,75°C ÷ 1,86°C/m.
b(solution) = 1,478 m = 1,478 mol/kg.