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aev [14]
2 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 -4.3°C?
_______ g
Chemistry
1 answer:
4vir4ik [10]2 years ago
7 0

The mass of the ethylene glycol used is 14.26 g

<h3>What is the freezing point?</h3>

The freezing point is the temperature at which a liquid is converted to a solid.

Given that;

ΔT = 0.0°C - ( -4.3°C) = 4.3°C

ΔT = 4.3°C

But we know that

ΔT = k m i

m = ΔT/k i

m = 4.3°C/1.86 °C/m * 1

m = 2.3 m

Now;

Let the mass of  ethylene glycol be m

Molar mass of  ethylene glycol = 62 g/mol

Mass of solvent = 100.0 g or 0.1 Kg

Then;

2.3= m/62/0.1

2.3 = m/62 * 0.1

m = 2.3 * 62 * 0.1

m = 14.26 g

Learn more about freezing point:brainly.com/question/3121416?

#SPJ1

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Nimfa-mama [501]
<span>Reaction of ethane combustion:

2C2H6 (g) + 7O2 (g) ----> 4CO2 (g) + 6H2O 

According to the reaction, we can see that </span>C2H6 and CO2 have following stoichiometric ratio: 

n(C2H6) : n(CO2) = 2 : 4

If we know the number of moles of ethane we can calculate moles of carbon dioxide:

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n(CO2) = 2 x n(C2H6)

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Since, in the reaction there are 2 moles of HCl,
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Explanation:

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