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3 years ago

How much energy is released when 50.00 mol of glycerin melts? (Heat of fusion glycerin-91.7 KJ/mol)

Chemistry

1 answer:

Irina-Kira [14]3 years ago

3 0

Answer:

4585 KJ

Explanation:

Data obtained from the question. This includes following:

Number of mole (n) of glycerin = 50 moles

Heat of fusion glycerin (Hf) = 91.7 KJ/mol

Heat (Q) released =?

The heat released can be obtained as follow:

Q = n•Hf

Q = 50 x 91.7

Q = 4585 KJ

Therefore, 4585 KJ of heat is released.

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What is the boiling point of an aqueous solution that has a vapor pressure of 23.0 torr at 25 ∘C? (P∘H2O=23.78 torr; Kb= 0.512 ∘

UNO [17]

Answer:

Boiling point of the solution is 100.964°C

Explanation:

In this problem, first, you must use Raoult's law to calculate molality of the solution. When you find the molality you can obtain the boiling point elevation because of the effect of the solute in the solution (Colligative properties).

Using Raoult's law:

Psol = Xwater × P°water.

As vapour pressure of the solution is 23.0torr and for the pure water is 23.78torr:

23.0torr= Xwater × 23.78torr.

0.9672 = Xwater.

The mole fraction of water is:

$0.9672 = \frac{X_{H_2O}}{X_{H_2O}+X_{solute}}$

Also,

$1 = X_{H_2O}+X_{solute}$

You can assume moles of water are 0.9672 and moles of solute are 1- 0.9672 = 0.0328 moles

Molality is defined as the ratio between moles of solute (0.0328moles) and kg of solvent. kg of solvent are:

$09672mol *\frac{18.01g}{1mol}* \frac{1kg}{1000g} = 0.01742kg$

Molality of the solution is:

0.0328mol Solute / 0.01742kg = 1.883m

Boiling point elevation formula is:

ΔT = Kb×m×i

<em>Where ΔT is how many °C increase the boiling point regard to pure solvent, Kb is a constant (0.512°C/m for water), m molality (1.883m) and i is Van't Hoff factor (Assuming a i=1).</em>

Replacing:

ΔT = 0.512°C/m×1.882m×1

ΔT = 0.964°C

As the boiling point of water is 100°C,

<h3>Boiling point of the solution is 100.964°C</h3>

<em />

5 0

3 years ago

Find the molecular weight of CO

Airida [17]

Answer:

The molecular weight of CO is 28.01g/mole.

6 0

3 years ago

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Find degenerate orbitals 3dxy, 2px, 4dz2, 3dyz, 3pz,3dxz, 4sm

IceJOKER [234]

Ok boomer..................

7 0

3 years ago

Look at the diagram of the atom. What will it change to during beta decay? A)An atom with 3 protons, 4 neutrons, and 3 electrons

GREYUIT [131]

D) An atom with 4 protons, 3 neutrons, and 3 electrons

5 0

4 years ago

a) Calculatethe molality, m, of an aqueous solution of 1.22 M sucrose, C12H22O11. The density of the solution is 1.12 g/mL.b) Wh

Contact [7]

Answer:

a) 1,74 molal

b) 37,2 %

c) 0,03

Explanation:

We are going to define sucrose as solute, water as solvent and the mix of both, the solution.

Let´s start with the data:

$Molarity = M = \frac{1,22 mol solute}{lts solution}$

We can assume as a calculus base, 1 liter of solution. So, in 1 liter of solution we have 1,22 moles of solute:

$1 lts solution * \frac{1,22 moles solute}{lts solution}=1,22 moles solute$

Knowing that the molality (m) is defined as mol of solute/kgs solvent, we have to calculate the mass of solvent on the solution. Remember our calculus base (1 lts of solution). In 1 lts of solution we have 1120 grams of solution.

$1 lts solution * \frac{1,12 grs solution}{mL solution}*\frac{1000 mL solution}{1 lts solution} = 1120 grs of solution$

With the molecular weight of solute (<em>Sum of: for carbon = 12*12=144; for hydrogen = 1*22=22 and for oxygen = 16*11=176. Final result = 342 grs per mol</em>), we can obtain the mass of solute:

$1,22 mol solute*\frac{342 grs solute}{1 mol solute} = 417,24 grs solute$

Now, the mass of solvent is: mass solvent = mass of solution - mass of solute. So, we have: 1120 - 417,24 = 702,76 grs of solvent = 0,70276 Kgs of solvent

$molality = m = \frac{1,22 mol solute}{0,70276 kgs solvent}= 1,74 molal$

For b) question we have that the mass percent of solute is hte ratio between the mass of solute and the mass of solution. So,

$%(w/w) = \frac{417,24 grs solute}{1120 grs solution} = 37,2%$

For c) question we have that the mole fraction of solute is the ratio between moles of solute and moles of solution. Let's calculate the moles of solution as follows: <em>Moles solution = moles solute + moles solvent.</em> First we have that the moles of solvent are (remember that the molecular weight of water for this calculus is 18 grs per mol):

$702,76 grs solvent*\frac{1 mol solvent}{18 grs solvent} = 39,04 moles solvent$

So, we have the moles of solution: 1,22 moles of solute + 39,04 moles of solvent = 40,26 moles of solution

Finally, we have:

$Mol frac solute = \frac{1,22 mol solute}{40,26 mol solution}= 0,03$

6 0

3 years ago