What would be the freezing point of a solution that has a molality of 1.324 m which was prepared by dissolving biphenyl (C12H10)
1 answer:
The freezing point of a 1.324 m solution, prepared by dissolving biphenyl into naphthalene, is 71.12 ° C.
A solution is prepared by dissolving biphenyl into naphthalene. We can calculate the freezing point depression (ΔT) for naphthalene using the following expression.
where,
- i: van 't Hoff factor (1 for non-electrolytes)
- Kf: cryoscopic constant
- m: molality
The normal freezing point of naphthalene is 80.26 °C. The freezing point of the solution is:
The freezing point of a 1.324 m solution, prepared by dissolving biphenyl into naphthalene, is 71.12 ° C.
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Heat is a type of energy that causes a person's body to feel hot or cold.
While the temperature of an object is a parameter that indicates how hot or cold the object is.
How is the temperature in degree Fahrenheit converted to degree celsius?
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Answer : The mass of ammonia produced can be, 121.429 k
Solution : Given,
Mass of = 100 kg = 100000 g
Mass of = 100 kg = 100000 g
Molar mass of = 28 g/mole
Molar mass of = 2 g/mole
Molar mass of = 17 g/mole
First we have to calculate the moles of and
.
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
From the balanced reaction we conclude that
As, 1 mole of react with 3 mole of
So, 3571.43 moles of react with
moles of
From this we conclude that, is an excess reagent because the given moles are greater than the required moles and
is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of
From the reaction, we conclude that
As, 1 mole of react to give 2 mole of
So, 3571.43 moles of react to give
moles of
Now we have to calculate the mass of
Therefore, the mass of ammonia produced can be, 121.429 kg
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