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Lorico [155]
3 years ago
10

The freezing point of pure water is 0.0°C. How many grams of ethylene glycol (C2H6O2) must be mixed in 100.0 g of water to lower

the freezing point of the solution to -8.8°C? g
Chemistry
1 answer:
SSSSS [86.1K]3 years ago
8 0

Answer:

17.54 g

Explanation:

the freezing point of the solution to -8.8°C

depression in freezing point = 8.8°C

Molal freezing point depression constant of ethylene glycol  Kf = 3.11⁰C /m

ΔTf = Kf x m , m is no of moles of solute per kg of solvent .

Let gram of ethylene glycol required be x .

moles = m / mol weight of ethylene glycol

= x / 62

100 g of water = .1 kg

moles of solute dissolved in 1 kg of water

m = x / 62 x .1

m = 10 x / 62

Using the above equation for depression in freezing point

8.8 = 3.11 x 10 x / 62

x = 17.54 g .

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(6.023x10^23)X = 4.91x10^22

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Answer:

d. each of the above (A, B, and C) occurs.

Explanation:

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Explanation:

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vazorg [7]

<u>Answer:</u> The correct answer is Option c.

<u>Explanation:</u>

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Liquid\rightleftharpoons Gas

The value of standard Gibbs free energy is 0 for equilibrium reactions.

To calculate \Delta S^o_{vap} for the reaction, we use the equation:

\Delta S^o_{vap}=\frac{\Delta H^o_{vap}}{T}

where,

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T = temperature of the reaction = 353.3 K

Putting values in above equation, we get:

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good luck!
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