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2 years ago

The density of a gaseous chlorofluorocarbon (CFC) at 23.8 °C and 432 mmHg is 3.23 g/L. What is its molar mass?

Chemistry

1 answer:

Wewaii [24]2 years ago

6 0

Answer:

138.57 g/mol.

Explanation:

The following data were obtained from the question:

Temperature (T) = 23.8 °C

Pressure = 432 mmHg

Density (D) = 3.23 g/L

Next, we shall obtain an expression for the density in relation to molar mass, pressure and temperature.

This can be obtained by using the ideal gas equation as shown below:

PV = nRT.... (1)

Recall:

Mole (n) = maas(m) /Molar mass (M)

n = m/M

Substituting the value of n into equation 1

PV = nRT

PV = mRT/M

Divide both side by P

V = mRT/MP

Divide both side by m

V/m = RT/MP

Invert the above equation

m/V = MP /RT..... (2)

Recall:

Density (D) = mass(m) /volume (V)

D = m/V

Replace m/V with D in equation 2

m/V = MP /RT

D = MP /RT

Thus, with the above formula we can obtain the molar mass of chlorofluorocarbon (CFC) as shown below:

Temperature (T) = 23.8 °C = 23.8 °C + 273 = 296.8 K

Pressure = 432 mmHg = 432/760 = 0.568 atm

Density (D) = 3.23 g/L

Gas constant (R) = 0.0821 atm.L/Kmol

Molar mass (M) =..?

D = MP /RT

3.23 = M x 0.568 / 0.0821 x 296.8

Cross multiply

M x 0.568 = 3.23 x 0.0821 x 296.8

Divide both side by 0.568

M = (3.23 x 0.0821 x 296.8) / 0.568

M = 138.57 g/mol

Therefore the molar mass of CFC is 138.57 g/mol.

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25. In the investigation shown the changes in heat of the copper is greater than the change in the heat of the water. What error

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Answer:

Heat lost to the surroundings

Heat lost to the thermometer

Explanation:

All changes in heat, or energy, can be explained. Many of the reactions or changes we see in the world involve the conversion of energy. For example as we heat up a substance (eg. water), the amount of energy we put in should give us an exact temperature. However, this is a "perfect world" scenario, and does not occur in real life. Whenever heat is added to a substance like water, we always need to account for the energy that is going to be lost. For example, heat lost to evaporation or even the effect of measuring the temperature with a thermometer (the introduction of anything including a thermometer will affect the temperature).

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2 years ago

A mysterious white powder could be powdered sugar (C12H22O11), cocaine (C17H21NO4), codeine (C18H21NO3), norfenefrine (C8H11NO2)

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Norfenefrine (C₈H₁₁NO₂).

<h3>Further explanation</h3>

We will solve a case related to one of the colligative properties, namely freezing point depression.

The freezing point of the solution is the temperature at which the solution begins to freeze. The difference between the freezing point of the solvent and the freezing point of the solution is called freezing point depression.

$\boxed{ \ \Delta T_f = T_f(solvent) - T_f(solution) \ } \rightarrow \boxed{ \ \Delta T_f = K_f \times molality \ }$

<u>Given:</u>

A mysterious white powder could be,

powdered sugar (C₁₂H₂₂O₁₁) with a molar mass of 342.30 g/moles,
cocaine (C₁₇H₂₁NO₄) with a molar mass of 303.35 g/moles,
codeine (C₁₈H₂₁NO₃) with a molar mass of 299.36 g/moles,
norfenefrine (C₈H₁₁NO₂) with a molar mass of 153.18 g/moles, or
fructose (C₆H₁₂O₆) with a molar mass of 180.16 g/moles.

When 82 mg of the powder is dissolved in 1.50 mL of ethanol (density = 0.789 g/cm³, normal freezing point −114.6°C, Kf = 1.99°C/m), the freezing point is lowered to −115.5°C.

<u>Question: </u>What is the identity of the white powder?

<u>The Process:</u>

Let us identify the solute, the solvent, initial, and final temperatures.

The solute = the powder
The solvent = ethanol
The freezing point of the solvent = −114.6°C
The freezing point of the solution = −115.5°C

Prepare masses of solutes and solvents.

Mass of solute = 82 mg = 0.082 g
Mass of solvent = density x volume, i.e., $\boxed{ \ 0.789 \ \frac{g}{cm^3} \times 1.50 \ cm^3 = 1.1835 \ g = 0.00118 \ kg \ }$

We must prepare the solvent mass unit in kg because the unit of molality is the mole of the solute divided by the mass of the solvent in kg.

The molality formula is as follows:

$\boxed{ \ m = \frac{moles \ of \ solute}{kg \ of \ solvent} \ } \rightarrow \boxed{ \ m = \frac{mass \ of \ solute \ (g)}{molar \ mass \ of \ solute \times kg \ of \ solvent} \ }$

Now we combine it with the formula of freezing point depression.

$\boxed{ \ \Delta T_f = K_f \times \frac{mass \ of \ solute \ (g)}{molar \ mass \ of \ solute \times kg \ of \ solvent} \ }$

It is clear that we will determine the molar mass of the solute (denoted by Mr).

We enter all data into the formula.

$\boxed{ \ -114.6^0C - (-115.5^0C) = 1.99 \frac{^0C}{m} \times \frac{0.082 \ g}{Mr \times 0.00118 \ kg} \ }$

$\boxed{ \ 0.9 = \frac{1.99 \times 0.082}{Mr \times 0.00118} \ }$

$\boxed{ \ Mr = \frac{0.16318}{0.9 \times 0.00118} \ }$

We get $\boxed{ \ Mr = 153.65 \ }$

These results are very close to the molar mass of norfenefrine which is 153.18 g/mol. Thus the white powder is norfenefrine.

<h3>Learn more</h3>

The molality and mole fraction of water brainly.com/question/10861444
About the mass and density of ethylene glycol as an antifreeze brainly.com/question/4053884
About the solution as a homogeneous mixture brainly.com/question/637791

Keywords: a mysterious white powder, sugar, cocaine, codeine, norfenefrine, fructose, the solute, the solvent, dissolved, ethanol, normal freezing point, the freezing point depression, the identity

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