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marysya [2.9K]
3 years ago
10

For a 0.50 molal solution of sucrose in water, calculate the freezing point and the boiling point of the solution.

Chemistry
1 answer:
jok3333 [9.3K]3 years ago
6 0

Answer:

-0.93 °C; 100.26 °C  

Step-by-step explanation:

(a) Freezing point depression

The formula for the freezing point depression ΔT_f is

ΔT_f = iKf·b

i is the van’t Hoff factor: the number of moles of particles you get from a solute.

For sucrose,

    Sucrose (s)   ⟶   sucrose (aq)

1 mole sucrose ⟶ 1 mol particles     i = 1

ΔT_f = 1 × 1.86 × 0.50

ΔT_f = 0.93 °C

 T_f = T_f° - ΔT_f  

 T_f = 0.00 – 0.93  

 T_f = -0.93 °C

(b) Boiling point elevation

The formula for the boiling point elevation ΔTb is

ΔTb = iKb·b

ΔTb = 1 × 0.512 × 0.50

ΔTb = 0.256 °C

 Tb = Tb° + ΔTb  

 Tb = 100.00 + 0.256  

 Tb = 100.26 °C

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When Ni(OH)₂ starts precipitate :

Ksp of Ni(OH)₂ = [ Ni²⁺ ] [ OH²⁻ ]

5.5x10⁻¹⁶ = [ 0.18 ] [ OH²⁻ ]

[ OH²⁻ ] = 5.5x10⁻¹⁶ / 0.18

[ OH⁻ ] = 5.5 × 10⁻⁸ M

pOH = 7.2

therefore , pH = 14 - 7.2

                  pH = 6.8

Thus, Precipitation calculations with Ni²⁺ and Pb²⁺ a. Use the solubility product for Ni(OH)₂ (s) . the pH at which Ni(OH)₂ begins to precipitate from a 0.18 M Ni²⁺ solution. (Ksp Ni(OH)₂ = 5.5x10⁻¹⁶) is 6.8.

To learn more about pH here

brainly.com/question/15289741

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