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Tanzania [10]
2 years ago
8

The boiling point of ethanol (C₂H₅OH) is 78.5°C. What is the boiling point of a solution of 6.4 g of vanillin M = 152.14 g/mol)

in 50.0 g of ethanol (Kb of ethanol = 1.22°C/m)?
Chemistry
1 answer:
Gwar [14]2 years ago
6 0

The boiling point of a solution of 6.4 g of vanillin in 50.0g of ethanol is 77.47 degree C.

Change in boiling point of solution

ΔTb = (Kb)(m), where

ΔTb = change in temperature

Kb = elevation constant = 1.22 °C./m

m  = molality of the solution

Following the calculation of the moles of vanillin, the molality of vanillin in ethanol will be determined..

Moles of vanillin = 6.4g / 152.12g/mol = 0.042 mol

m of vanillin = 0.042mol / 0.05kg = 0.84m

putting all values on the above equation,

ΔTb = (1.22)(0.84) = 1.0248 °C.

Given;

boiling point of ethanol = 78.5 °C.

So the boiling point of vanillin = 78.5 - 1.0248 = 77.47 °C

Hence, the boiling point of a solution of vanillin is 77.47 °C.

Learn more about the boiling point with the help of the given link:

brainly.com/question/1514229

#SPJ4

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Here's what I get  

Explanation:

(i) Voltaic cell

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(ii) Overall Cell Potential

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                               <u>ℰ°/V </u>

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Zn²⁺ + 2e⁻ ⇌ Zn   -0.76

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The half-reaction with the more negative potential is the oxidation half-reaction. It is the  reaction that occurs at the anode.

We reverse that half-reaction and subtract the voltages to get the cell reaction.

                                                         <u> ℰ°/V </u>

Cathode:  Cu²⁺ + 2e⁻ ⇌ Cu              0.34

<u>Anode:     Zn ⇌ Zn²⁺ + 2e⁻              -0.76</u>

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\mathcal{E}_{\text{cell}}^{\circ} = \mathcal{E}_{\text{cat}}^{\circ} - \mathcal{E}_{\text{an}}^{\circ} = \text{0.34 V} - \text{(-0.76 V)} = \text{0.34 V} + \text{0.76 V} = \textbf{1.10 V}

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The specific labels will depend on your textbook.

They are often as follows.

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c. Electron flow

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f. Salt bridge

g. Zn²⁺(aq) and NO₃⁻(aq)

h. Anode or Zn

The salt bridge enables ions to flow in the internal circuit and to maintain electrical neutrality in the two compartments.

It often consists of a saturated solution of KCl.

As Zn²⁺ ions form in the anode compartment, Cl⁻ ions move in to provide partners for them.

As Cu²⁺ ions are removed from the cathode compartment, K⁺ ions move in to replace them.

 

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