Video of chemical material delivery about the nature of the colligative

Colligative nature

Link video: https://www.youtube.com/resultssearch_query=https%3A%2F%2Fyoutu.be%2F72aSUJssLpw

1. Colligative nature

Colligative nature is the physical properties of a solution that depends only on the concentration of solute particles, but not on the type. The electrolyte solution has a greater colligative property than the non-electrolyte solution having the same concentration because the electrolyte solution has a larger number of solute particles.

A. Kemolalan (m)

The molecularity or molality indicates the number of moles (n) of solute in 1 kg (= 1,000 g) solvent. Therefore, the kemolalan is expressed in mol kg-1.

M = n / p

M = concentration of solution

N = number of moles of solute

P = mass of solvent (in kg)

Example:

What is the solubility of the solution made by mixing 3 grams of urea with 200 grams of water?

Answer: a solution of 3 grams of urea in 200 grams of water.

Mol urea = 3/60 g mol-1 = 0.05 mol

The solvent mass = 200 grams = 0.2 kg

M = n / p = 0.05 mol = 0.25 mol kg-1

0.2 kg

B. Mole fraction (X)

The mole fraction (x) represents the ratio of the number of moles of solute or solvent to the number of moles of the solution. If the number of moles of solvent is nA, and the number of moles of solute is nB, then the mole fraction of the solvent and solute is:

The amount of solvent mole fraction with solute is 1

XA + XB = 1

Example:

Calculate the mole fraction of urea in a 20% urea solution (Mr. urea = 60)

Answer:

In 100 grams of 20% urea solution there are 20 grams and 80 grams of water.

Water mol = 80 g / 18 g mol-1 = 4.44 mol

Mol urea = 20 g / 60 g mol-1 = 0.33 mol

X urea = XB = 0.33 mol / (4.44 + 0.33) mol = 0.069

C. Solvent Pressure

The vapor pressure of a substance is the pressure generated by the saturated vapor of the substance. The higher the temperature, the greater the vapor pressure. If the solute does not evaporate then the vapor pressure of the solution becomes lower than the vapor pressure of the solvent. The difference between purified solvent vapor (P0) and the vapor pressure of the solution (P) is called the decrease in the vapor pressure of the solution (ΔP).

D. Raise the Boiling Point

The solution has a higher boiling point and a freezing point lower than the solvent. The difference between the boiling point of the solution and the boiling point of the solvent is called the boiling point increase (ΔTb). Formula: ΔTb = Kb x m

Where: m = molality of solution

Kb = the increase in boiling point

E. Decrease of Freezing Point

Increase the boiling point and decrease of freezing point is proportional to the solubility of the solution: ΔTb = m x Kb and ΔTf = m x Kf. The difference between the freezing point of the solvent and the freezing point of the solution is called the freezing point drop (ΔTf). The boiling point increase and the decrease in freezing point of the solution can be explained by the phase diagram

F. Osmotic Pressure

• Osmosis is solvent molecular seepage from solvent into solution, or from a dilute solution to a more concentrated solution, via a semipermiable membrane.

• The osmotic pressure is the pressure that must be applied to the surface of the solution to prevent osmosis from pure solvent.

• Formula: л = M. R .T

• Solutions having the same osmotic pressure are called isotonic

Try To make LEARNING IMPLEMENTATION PLAN (RPP) with curriculum 2013

LEARNING IMPLEMENTATION PLAN (RPP)

Education Unit : SMA 2 TEBO

Class / Semester : X / 2

Subject : Chemistry

Principal Material : Compound Nomenclature Based on IUPAC

1st Meeting : 1

Allocation Time: 2 x 45 minutes

A. Core Competencies:

KI 1: Living and practicing the religious teachings it embraces

KI 2: Living and practicing honest, disciplined, responsible, caring (polite, cooperative, tolerant, peaceful) behavior, polite, responsive and pro-active and showing attitude as part of the solution to problems in interacting effectively with the social environment and Nature and in placing ourselves as a reflection of the nation in the association of the world

KI 3: Understanding, applying, analyzing factual, conceptual, procedural knowledge based on his knowledge of science, technology, art, culture, and humanities with the insights of humanity, nationality, state and civilization on the causes of phenomena and events, and applying procedural knowledge to A specific field of study according to his or her talents and interests to solve problems

KI 4: Processing, reasoning, and recruiting in the realm of concrete and abstract realms related to the development of the self-study in the school independently, and able to use methods according to scientific rules

B. Basic Competencies:

KD of KI 1: Recognizes order in all chemical compound interactions as a form of the power of God Almighty.

KD from KI 2: Demonstrate co-operative, courteous, tolerant, peace-loving and caring about the environment and cost-effective in utilizing natural resources.

KD of KI 3: Applying IUPAC rules for naming simple inorganic and organic compounds.

KD of KI 4: Reasoning the IUPAC rules for naming simple inorganic and organic compounds.

C. Indicators of Achievement:

Shows attitudes and attitudes of receiving, respecting, and carrying out honesty, precision, discipline and responsibility in explaining the nomenclature of simple inorganic and organic compounds according to IUPAC rules.

D. Learning Objectives:

Students are able to demonstrate attitudes of behavior and attitudes of receiving, respecting, and carrying out honesty, rigor, discipline and responsibility in explaining the nomenclature of simple inorganic and organic compounds according to IUPAC rules.

E. Learning Materials:

• The naming of binary compounds

• The nomenclature of polyatomic compounds

F. Learning Method:

• Approaches: Scientific

• Model: Kooferatif tife PBL

• Method: discussion, training, assignment

G. Learning Activities

Activites	Description	Time Allocation
preliminary	• Teachers deliver learning objectives • Teachers motivate students by asking questions related to the material • Students form groups on teacher guidance	15 minute
Core	Observe : • Review the literature on the nomenclature of simple inorganic and organic compounds according to IUPAC rules. Questioning: • Student Ask the question how to apply the IUPAC rules to name the compound. Reasoning : • Students review the literature to answer questions relating to the nomenclature of simple  inorganic and organic compounds according to IUPAC rules. • Students discuss the IUPAC rules to name the compound. Trying: • Students conclude the application of simple inorganic and organic compound rule rules according to IUPAC rules. • Students practice naming compounds according to IUPAC rules. Communicating: • Students present the application of simple inorganic and organic compound rule rules according to IUPAC rules using correct grammar	60 minute
cover	• Teacher gives reinforcement of material that has been taught • Teacher gives assignment for next meeting	15 minute

H. Learning Tools and Resources:

1. Tools and Materials

• Notebook

• LCD Projector

2. Learning Resources

• LKS

• Chemistry book class X

I. Assessment of Learning Processes and Outcomes:

Assessment Sheet Specification Table

Aspects/Indicators	Techniques	Forms	Instruments (tes dan non tes)
Affective(KI 1 & KI 2)	Observation	Observation	Observation sheet (terlampir)
Knowledge (KI 3)	Written test	Description	Description problems(terlampir)
Skil  (K4)

Appendix 1: Attitudinal Attitude Observation Format

Name	Perilaku yang diamati dalam pembelajaran
	Honest	Discipline	Responbility	caring	Hard work	Respect for others

Information: Attitude competency indicator

Honest A. Convey something based on the actual situation B. Does not cover the errors that occur 2. Discipline A. Always present in class on time B. Work on LKS as directed and timely C. Obey the rules of the game in independent and group work 3. Responsibility A. Trying to complete the task seriously B. Ask friends / teachers when you encounter problems C. Resolving the issues that are his responsibility D. Participation in groups 4. Caring A. Keeping the class clean, helping the friends in need B. Show empathy and sympathy to solve problems C. Able to give an idea / idea to an existing problem in the vicinity D. Provide assistance in accordance with his abilities 5. Work hard A. Doing LKS seriously B. Show unyielding attitude C. Trying to find a solution to the given problem 6. Respect others A. Appreciate the opinions of friends or other groups B. Show calm to listen to teacher explanation Scoring scale : A = very good B = good C = less D = very less Appendix 2: Knowledge Assessment Instrument Problem Description: 1. Give the name of the following compound: A. Na2O B. Mg (OH) 2 2. Write the chemical formula of the following compounds: A. Dinitrogen pentaoxide B. Iron (III) oxide

Using English to Predict Rendement of Product Reaction

PREDICT RENDEMENT OF PRODUCT OF REACTION




Rendement is the percentage of comparison between the initial weights generated

With some final weight gain Rendement is The percentage of material that can be obtained, for example a ratio between the net proceeds and the gross proceeds the chemical yield, the yield of

the reaction, or only the rendement refers to the amount of reaction product

produced in the chemical reaction. Absolute rendement can be written as weight

in grams or in moles (molar yield). The relative yield used as a calculation of
the effectiveness of the procedure, is calculated by dividing the amount of

ement = actual yield of theoretical rendement / initial yield multiplie
product obtained in moles by the theoretical yield in moles: Fractional
ren
dd by
100% To obtain a percentage yield, multiply the fractional yield by 100%. One


rendement is calculated based on the number of moles of the

limiting re

or more reactants in chemical reactions are often used redundantly. The
theoretical
 agent. For this calculation, it is usually assumed that there is
only one reaction involved. The ideal chemical yield value (theoretical


owing equations

percent rendemen = weight yield / weight of yield divided b

rendement) is 100%, a value highly unlikely to be achieved in its practice.
Calculate the percentage of rendemen that is by using the fol
ly the sample weight


multiplied by 100%

Fractional yield = the actual rendement/theoritical rendement x 100%

• Water Hydration Hydrate compounds are related compounds or molecules in hydra.t while anhydrous compounds are compounds that lose water molecules due to continuous heating. Hydrate compounds are also called crystalline compounds containing molecules that have hydrogen bonds. Water is a versatile compound that participates in various chemical reactions on earth. Understanding of water hydration is very beneficial for every aspect of life. In the pharmaceutical field, the hydration principle of water is used in the manufacture of alcohols through the direct hydration of alkenes (Ibn, 2007: 11). Some of the reactions done in the kimi laboratory are always concerned with the solution, some of which use water as a solvent. When water is evaporated, the reaction product can be isolated, often in solid form. For example, if nickel (H) oxide (N, OL) is dissolved in dilute H2SO4 solution, NiSO4: NiO (s) + H2SO4 (ag) à NiSO4 (ag) + H2O (I)

When water is evaporated, a dark green crystal is formed. When analyzed the crystals contain 6 moles of water for each nickel mole (ii) Oxide. This compound is called hydrate or gram hydrate and water is an important part of the composition formed and is called water hydrate (Epinur, 20015: 31).

PRACTICAL METHOD 3.1 Tools and Materials

• Tools                                                                               

- Crucifix and cover plate

- Paper towels

- Segitia porseloen

- Bunsen burners

- Crusher brace

-Drop pipette

-Tripod

- One

- Glass Watch

- Balance Sheet

- Segitia Porselen

• Material - Magnesium Band 10-15 cm - Distilled water - 0.5 gram copper metal - Detergent - 6M HNO3 solution - Sample 1 gram - Nitrate acid 4 M 10 ml - Copper (II) sulfate pentahydrate (CuSO4,5H20) ½ spatula

 • Water hydration A. Quantitative determination of water percentage in hydrate compounds Porcelain Cup + Cover Washed detergent and water and rinsed with distilled water + 6M HNO3 solution Rinsed with distilled water once more  And placed on a buffer triangle The height of the three legs is arranged, so that the hands of the cup are right on the hot part The cover is slightly opened when heated Heated until the cup burns up to 5 minutes The heating is stopped and cooled to room temperature for 10-15 minutes

Cup + sample + cover

Weighed

Put on triangle and cover slightly opened so that steam can come out

Heated for 1 minute

Heat is raised until the top of the cup looks mereah

The heating is left for 10 minutes

The heating is stopped

Closed cup

Cooled at room temperature

• Water

1. Mass of empty saucer + cover 128,70 gram

2. Mass of the bowl + lid + example 129.70 grams

3. Empty bowl masses + lid + sample 129.52 grams

(Heating 1)

4. Cawankosong mass + close + example 129,43 gram

(Heating 2)

5. Mask empty bowl + cover + example 129.44 grams

6. Example mass after heating (0.74 gram fixed weight)

7. Mass sample after heating 0.74 grams

8. The lost water mass of the sample is 0.26 gram

9. the percentage of water lost from the sample 26%

10. The molar mass of the adhydrate compound is 4.6 x 10-3 mol

11. Hydrate formula CuSO4, 5H2O

12. Number of substance anu 0,26 gram

• Water Hydration Hydrates are compounds that still contain water. Some of the reactions done in the chemistry laboratory are always concerned with using water as a solvent. When water is evaporated, the reaction product can be isolated, often in solid form. Sometimes these solids contain water molecules as part of their composition. This experiment aims to study the properties of the dihydrate compound, the alternating reaction and the percentage of water in a hydrate. The ingredients used for this lab are the example substances of the assistant CuSO4 5H2O. CuSO4 binds some water molecules. In the experiment hydrate compounds will become anhydrous compounds because crystalline water is removed by heating to obtain inhydrate salts characterized by changes in color, form and container where heating will dry from the water molecule. Through the heating process hydrate compounds will be released. The tools required in this lab is a porcelain cup and a lid in the form of a watch glass. Used watch glass because of limited tools and water vapor can attach to the glass and can be seen. The cup is cleaned, washed with distilled water and dried. For cleaning, it should use HNO3, however, the HNO3 solution is very dangerous. Therefore, just simply rinse with distilled water. The cup is placed on the three legs and gauze. The cup is heated with a slightly opened cover. Warming done for 10-15 minutes. The heating function of the tiatu to evaporate the water contained in the substance. Then cooled at room temperature about 10 minutes. The cup and the lid were weighed and weighed 128.70 gra. The sample substance (CuSO4.5H2O) was obtained from the assistant approximately 1 gram, placed on the cwan and weighed again, the weight being 129.70 grams. The cup + sample + lid is heated with bunsen and the lid is opened slightly about 10 minutes of heating. The hearth was stopped, the cwan was closed, and cooled at room temperature, then weighed 129.52 grams. In the second warming result obtained 129.43 grams and on the third heating result 129.44 grams. Example mass after heating can be obtained water of difference between (mass of cup + lid + sample + heating 3) with (empty plate mass + lid) that is 0.74 gram. The mass of water lost from the sample can be from the difference before heating to after heating 3 is 0.26 gram. The percentage of lost water can be searched by the formula of yield:  

the mas of water lost/early water mass _{X 100%}

The result is 20%. The molar period of anhydrous compounds At the time of warming the color change becomes faded, it signifies the loss of water and evaporates The weight after heating 3 times also decreases and the water evaporates about 26% CuSO45H2O means, mole of copper (II) sulfate binds 5 water molecules. 1. The percentage of water in a hydrate can be determined by the formula: % Water =

the mas of water lost/early water mass _{X 100%}

• Water Hydration

1. Mass of empty saucer + cover 128,70 gram

2. Mass of the bowl + lid + example 129.70 grams

3. Empty bowl masses + lid + sample 129.52 grams

(Heating 1)

4. Cawankosong mass + close + example 129,43 gram

(Heating 2)

5. Mask empty bowl + cover + example 129.44 grams

(Heating 3)

6. Mass sample after heating (fixed weight)

= (Mass of cup + lid + sample of heating 3) - (Mass of cup + lid)

= 129.44 - 128.70 grams

= 0.74 grams

7. The lost water mass of the sample

= (Mass of cup + lid + sample) - (Mass of cup + lid + sample processing) = 129.70 - 129.44 grams

= 0.26 grams

8. the percentage of water lost from the sample

% Water = the mas of water lost/early water mass _{X 100%}

=1/26 x100%

= 26%

9. Number of substances: 0.26 grams