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2 years ago

A 0.050 M solution of AlCl3 had an observed osmotic pressure of 3.85 atm at 20 ∘C . Calculate the van't Hoff factor i for AlCl3.

Chemistry

1 answer:

saw5 [17]2 years ago

4 0

Answer:

Van't Hoff factor, for this solution is 3.2

Explanation:

Osmotic pressure formula → π = M . R . T . i

π = Pressure → 3.85 atm

M = Molarity → 0.05mol/L

R = Constant for the Ideal Gases → 0.082 L.atm / mol.K

T = Absolute T° (K) → T°C + 273

20°C + 273 = 293K

Let's replace data:

3.85 atm = 0.05 mol/L . 0.082 L.atm/mol.K . 293K . i

3.85 atm / 0.05 . 0.082 atm . 293 = i

3.20 = i

If we assume complete ionization, i must be ≅ 4

AlCl₃ → Al³⁺ + 3Cl⁻

We have 1 mol of aluminum cation and 3 moles of chloride, 4 total moles.

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Describe the many different forms of energy involved with stretching and releasing a rubber band. What other processes are simil

Nadya [2.5K]

Answer:

Conversion of kinetic energy to potential energy (chemo mechanical energy)

In the state of rest, the rubber is a tangled mass of long chained cross-linked polymer that due to their disorderliness are in a state of increased entropy. By pulling on the polymer, the applied kinetic energy stretches the polymer into straight chains, giving them order and reducing their entropy. The stretched rubber then has energy stored in the form of chemo mechanical energy which is a form of potential energy

Conversion of the stored potential energy in the stretched to kinetic energy

By remaining in a stretched condition, the rubber is in a state of high potential energy, when the force holding the rubber in place is removed, due to the laws of thermodynamics, the polymers in the rubber curls back to their state of "random" tangled mass releasing the stored potential energy in the process and doing work such as moving items placed in the rubber's path of motion such as an object that has weight, w then takes up the kinetic energy 1/2×m×v² which can can result in the flight of the object.

Explanation:

5 0

1 year ago

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Which of the following statements best describes the properties of aluminum-28?

Contact [7]

d. When aluminum-28 undergoes beta decay, silicon-28 is produced.

Explanation:

When the nuclei of aluminium-28 decays, it produces silicon- 28:

Aluminium ²⁸₁₃Al

Silicon 28 ²⁸₁₄Si

beta particle ⁰₋₁ $\beta$

²⁸₁₃Al → ²⁸₁₄Si + ⁰₋₁ $\beta$

This way, the mass and atomic number are conserved.

Conservation of mass number:

28 = 28 + 0, 28 = 28

13 = 14 -1 , 13 = 13

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Balancing nuclear equations brainly.com/question/10094982

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2 years ago

100 POINTS PLEASE HELP!! Honors Stoichiometry Activity Worksheet Instructions: In this laboratory activity, you will taste test

Shtirlitz [24]

Answer:

2 water + sugar + lemon juice → 4 lemonade

Moles of water present in 946.36 g of water=\frac{946.36 g}{236.59 g/mol}=4 mol=

236.59g/mol

946.36g

=4mol

Moles of sugar present in 196.86 g of water=\frac{196.86 g}{225 g/mol}=0.8749 mol=

225g/mol

196.86g

=0.8749mol

Moles of lemon juice present in 193.37 g of water=\frac{193.37 g}{257.83 g/mol}=0.7499 mol=

257.83g/mol

193.37g

=0.7499mol

Moles of lemonade in 2050.25 g of water=\frac{2050.25 g}{719.42 g/mol}=2.8498 mol=

719.42g/mol

2050.25g

=2.8498mol

As we can see that number of moles of lemon juice are limited.

So, we will consider the reaction will complete in accordance with moles of lemon juice.

1 mole lemon juice reacts with 2 mol of water,then 0.7499 mol of lemon juice will react with:

\frac{2}{1}\times 0.7499 mol = 1.4998 mol

×0.7499mol=1.4998mol of water

Mass of water used = 1.4998 mol × 236.59 g/mol=354.8376 g

Water remained unused = 946.36 g - 354.8376 g =591.5223 g

1 mole lemon juice reacts with mol of sugar,then 0.7499 mol of lemon juice will react with:

\frac{1}{1}\times 0.7499 mol = 0.7499 mol

×0.7499mol=0.7499mol of water

Mass of sugar used = 0.7499 mol × 225 g/mol = 168.7275 g

Sugar remained unused = 196.86 g - 28.1325 g

1 mole of lemon juice gives 4 moles of lemonade.

Then 0.7499 mol of lemon juice will give:

\frac{4}{1}\times 0.7499 mol=2.996 mol

×0.7499mol=2.996mol of lemonade

Mass of lemonade obtained = 2.996 mol × 719.42 g/mol = 2157.9722 g

Theoretical yield of lemonade = 2157.9722 g

Experimental yield of lemonade = 2050.25 g

Percentage yield of lemonade:

\frac{\text{Experimental yield}}{\text{theoretical yield}}\times 100

theoretical yield

Experimental yield

×100

\frac{2050.25 g}{2157.9722 g}\times 100=95.00\%

2157.9722g

2050.25g

×100=95.00%

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2 years ago

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The atoms of a certain element each contain 19 protons and 1 valence electron. Which statement correctly identifies this element

Leya [2.2K]

Answer

its a potassium elements

Explanation:

K 2,8,8,1 period num 4

group 1A

or 1s1 ,2s2 2p6, 3s1

its a metal reacted rapidly with water to form a colorless basic solution of potassium hydroxide (KOH) and hydrogen gas (H2). The reaction continues even when the solution becomes basic. The resulting solution is basic because of the dissolved hydroxide. The reaction is exothermic.

3 0

1 year ago

A 15.0 mL sample of 0.013 M HNO3 is titrated with 0.017 M CH$NH2 which he Kb=3.9 X 10-10. Determine the pH at these points: At t

kramer

Answer: The pH of the solution in the beginning is 1.89 and the pH of the solution after the addition of base is

Explanation:

For 1: At the beginning

To calculate the pH of the solution, we use the equation:

$pH=-\log[H^+]$

We are given:

Nitric acid is a monoprotic acid and it dissociates 1 mole of hydrogen ions. So, the concentration of hydrogen ions is 0.013 M

$[H^+]=0.013M$

Putting values in above equation, we get:

$pH=-\log(0.013)\\\\pH=1.89$

For 2:

To calculate the number of moles, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$

For nitric acid:

Molarity of nitric acid solution = 0.013 M

Volume of solution = 15 mL

Putting values in above equation, we get:

$0.013M=\frac{\text{Moles of }HNO_3\times 1000}{15}\\\\\text{Moles of }HNO_3=1.95\times 10^{-4}mol$

For methylamine:

Molarity of methylamine solution = 0.017 M

Volume of solution = 10 mL

Putting values in above equation, we get:

$0.017M=\frac{\text{Moles of }CH_3NH_2\times 1000}{10}\\\\\text{Moles of }CH_3NH_2=1.7\times 10^{-4}mol$

The chemical equation for the reaction of nitric acid and methylamine follows:

$HNO_3+CH_3NH_2\rightarrow CH_3NH_3^++NO_3^-$

As, the mole ratio of nitric acid and methyl amine is 1 : 1. So, the limiting reagent will be the reactant whose number of moles are less, which is methyl amine.

By Stoichiometry of the reaction:

1 mole of methyl amine produces 1 mole of $CH_3NH_3^+$