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Lubov Fominskaja [6]
3 years ago
15

What is the percent composition of iron in iron III chloride?

Chemistry
1 answer:
lara31 [8.8K]3 years ago
7 0
The correct answer is c
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The osmotic pressure of a solution containing 2.04 g of an unknown compound dissolved in 175.0 mLof solution at 25 ∘C is 2.13 at
kherson [118]

<u>Answer:</u> The molecular formula of the compound is C_4H_{10}O_4

<u>Explanation:</u>

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

\pi=iMRT

Or,

\pi=i\times \frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}\times RT

where,

\pi = osmotic pressure of the solution = 2.13 atm

i = Van't hoff factor = 1 (for non-electrolytes)

Given mass of compound = 2.04 g

Volume of solution = 175.0 mL

R = Gas constant = 0.0821\text{ L atm }mol^{-1}K^{-1}

T = temperature of the solution = 25^oC=[273+25]=298K

Putting values in above equation, we get:

2.13atm=1\times \frac{2.04\times 1000}{\text{Molar mass of compound}\times 175.0}\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 298K\\\\\text{Molar mass of compound}=\frac{1\times 2.04\times 1000\times 0.0821\times 298}{2.13\times 175.0}=133.9g/mol

  • <u>Calculating the molecular formula:</u>

The chemical equation for the combustion of compound having carbon, hydrogen and oxygen follows:

C_xH_yO_z+O_2\rightarrow CO_2+H_2O

where, 'x', 'y' and 'z' are the subscripts of carbon, hydrogen and oxygen respectively.

We are given:

Mass of CO_2=36.26g

Mass of H_2O=14.85g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

<u>For calculating the mass of carbon:</u>

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 36.26 g of carbon dioxide, \frac{12}{44}\times 36.26=9.89g of carbon will be contained.

<u>For calculating the mass of hydrogen:</u>

In 18 g of water, 2 g of hydrogen is contained.

So, in 14.85 g of water, \frac{2}{18}\times 14.85=1.65g of hydrogen will be contained.

Mass of oxygen in the compound = (22.08) - (9.89 + 1.65) = 10.54 g

To formulate the empirical formula, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon = \frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{9.89g}{12g/mole}=0.824moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1.65g}{1g/mole}=1.65moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{10.54g}{16g/mole}=0.659moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.659 moles.

For Carbon = \frac{0.824}{0.659}=1.25\approx 1

For Hydrogen = \frac{1.65}{0.659}=2.5

For Oxygen = \frac{0.659}{0.659}=1

Converting the mole fraction into whole number by multiplying the mole fraction by '2'

Mole fraction of carbon = (1 × 2) = 2

Mole fraction of oxygen = (2.5 × 2) = 5

Mole fraction of hydrogen = (1 × 2) = 2

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 2 : 5 : 2

The empirical formula for the given compound is C_2H_5O_2

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is:

n=\frac{\text{Molecular mass}}{\text{Empirical mass}}

We are given:

Mass of molecular formula = 133.9 g/mol

Mass of empirical formula = 61 g/mol

Putting values in above equation, we get:

n=\frac{133.9g/mol}{61g/mol}=2

Multiplying this valency by the subscript of every element of empirical formula, we get:

C_{(2\times 2)}H_{(5\times 2)}O_{(2\times 2)}=C_4H_{10}O_4

Hence, the molecular formula of the compound is C_4H_{10}O_4

4 0
3 years ago
A gas is heated from 263.0 K to 298.0 K and the volume is increased from 24.0 liters to 35.0 liters by moving a large piston wit
Evgesh-ka [11]

Answer:

The answer to your question is   P2 = 0.78 atm

Explanation:

Data

Temperature 1 = T1 = 263°K                 Temperature 2 = T2 = 298°K

Volume 1 = V1 = 24 L                             Volume 2 = V2 = 35 L

Pressure 1 = P1 = 1                                  Pressure 2 = P2 = ?

Process

1.- To solve this problem use the Combined gas law

                          P1V1/T1 = P2V2/T2

-Solve for P2

                           P2 = P1V1T2 / T1V2

-Substitution

                          P2 = (1)(24)(298) / (263)(35)

-Simplification

                          P2 = 7152 / 9205

-Result

                          P2 = 0.777

   or                    P2 = 0.78 atm

5 0
3 years ago
An aqueous solution contains 32.7% KCl (wt/wt%). what is the mole fraction of KCl in the solution
inna [77]

The mole fraction of KCl in the solution is 0.1051

calculation

mole fraction of KCl in solution = moles of KCl / total number of moles(moles of KCl +moles of H2O)

moles=mass/molar mass

mass of KCl=32.7g

molar mass of KCl= 39 +35.5

moles of KCl is therefore= 32.7g/74.5 g/mol=0.439 moles

find the moles of H2O= mass of H2O/molar mass

mass of H2O=100-32.7=67.3g

molar mass of H2O=( 1 x2) +16=18 g/mol

moles = 67.3/18 =3.739 moles

total moles=3.739+0.439=4.178 moles

mole fraction is therefore=0.439/4.178=0.1051

5 0
3 years ago
ASAP worth 15 points also
Taya2010 [7]
The person above me is correct I took a test on this so it’s the right answer
7 0
3 years ago
Please help on the questions that is left blank!! Due soon and will mark brainliest
34kurt

Answer: The strength of an acid or alkali depends on the degree of dissociation of the acid or alkali in water. The degree of dissociation measures the percentage of acid molecules that ionise when dissolved in water. He could use universal indicators or litmus paper for this.

i hope this helps you!

3 0
3 years ago
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