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Calculate the freezing point of a 0.08500 m aqueous solution of nano3. the molal freezing-point-depression constant of water is 1.86°c/m. remember to include the value of i.

Respuesta :

PBCHEM
Depression in freezing point (Δ[tex] T_{f} [/tex]) = [tex] K_{f} [/tex]×m×i,
where, [tex] K_{f} [/tex] = cryoscopic constant = [tex] 1.86^{0} C/m[/tex],
m= molality of solution = 0.0085 m
i = van't Hoff factor = 2 (For [tex] NaNO_{3} [/tex])

Thus, (Δ[tex] T_{f} [/tex]) = 1.86 X 0.0085 X 2 = [tex] 0.03162^{0}C [/tex]

Now, (Δ[tex] T_{f} [/tex]) = [tex] T^{0} [/tex] - T
Here, T = freezing point of solution
[tex] T^{0} [/tex] = freezing point of solvent = [tex] 0^{0}C [/tex]
Thus, T = [tex] T^{0} [/tex] - (Δ[tex] T_{f} [/tex]) = -[tex] 0.03162^{0}C [/tex]

Answer : The freezing point of a solution is, [tex]0.32^oC[/tex]

Explanation :

First we have to calculate the Van't Hoff factor (i) for [tex]NaNO_3[/tex].

The dissociation of [tex]NaNO_3[/tex] will be,

[tex]NaNO_3\rightarrow Na^++NO_3^-[/tex]

So, Van't Hoff factor = Number of solute particles = [tex]Na^++NO_3^-[/tex] = 1 + 1 = 2

Now we have to calculate the freezing point of a solution.

Formula used for lowering in freezing point :

[tex]\Delta T_f=i\times k_f\times m[/tex]

or,

[tex]T_f^o-T_f=i\times k_f\times m[/tex]

where,

[tex]\Delta T_f[/tex] = change in freezing point

[tex]T_f[/tex] = temperature of solution = ?

[tex]T^o_f[/tex] = temperature of pure water = [tex]0^oC[/tex]

[tex]k_f[/tex] = freezing point constant  = [tex]1.86^oC/m[/tex]

m = molality  = 0.08500 m

i = Van't Hoff factor = 2

Now put all the given values in this formula, we get the freezing point of a solution.

[tex]0^oC-T_f=2\times (1.86^oC/m)\times 0.08500m[/tex]

[tex]T_f=0.32^oC[/tex]

Therefore, the freezing point of a solution is, [tex]0.32^oC[/tex]

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