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 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.

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OlaMacgregor OlaMacgregor · Answer 2 · 2019-11-23T17:46:43+01:00

Explanation:

Relation between freezing temperature and molal concentration is as follows.

[tex]\Delta T_{f} = k_{f} \times m[/tex]

The given data is as follows.

[tex]\Delta T_{f}[/tex] = difference in temperature = [tex][0 - (-2.75)]^{o}C[/tex] = [tex]2.75^{o}C[/tex]

[tex]k_{f} = 1.86^{o}C/mol[/tex]

molality, (m) = ?

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

m = [tex]\frac{\Delta T_{f}}{k_{f}}[/tex]

= [tex]\frac{2.75^{o}C}{1.86^{o}C/mol}[/tex]

= 1.48 m

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