GENERAL CHEMISTRY 2

2nd Semester - Module 3 SOLUTION

CONCENTRATION AND STOICHIOMETRY

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Republic of the Philippines

Department of Education Regional Office IX, Zamboanga Peninsula

General Chemistry 2 – G11/12 Support Material for Independent Learning Engagement (SMILE) Module 3: Concentration and Solution Stoichiometry First Edition, 2021

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1

What I Need to Know

This module was designed and written to help you master the ways of expressing

the concentration of solutions and perform stoichiometric calculations involving

solutions. After going through this module, you are expected to:

1. Use different ways of expressing the concentration of solutions: percent by

mass, mole fraction, molarity, molality, percent by volume, percent by mass, ppm

(STEM_GC11PPIIId-f-111); and

2. Perform stoichiometric calculations for reactions in solution

(STEM_GC11PPIIId-f-112).

3. Describe laboratory procedures in determining the concentration of solutions

(STEM_GC11PPIIId-f-119)

What’s In

Activity 1: Solution Vocabulary Direction: Form the scrambled letters into words that are associated with a solution. Refer to the words listed below.

1. TNACROOITNENC _____________________________________ 2. TNVELOS _____________________________________ 3. ANERDNCTTOCE _____________________________________ 4. ISLOTNUO _____________________________________ 5. TULSOBILIY _____________________________________ 6. ITULDE _____________________________________ 7. TIVUIANETQTA _____________________________________

What’s New

Activity 2: Just Concentrate

Direction: Complete the table below by putting the correct information from the given solution concentration.

Solution

Concentration

Amount of

Solute

Amount of

Solvent

Amount of

Solution

10% (wt/wt) 10 g 90 g 100 g

20% by volume 20 ml 3. 100 ml

5% mass/volume 5 g - 5.

0.85 molal 0.85 mol 4. -

0.350 molar 1. - 1 L

0.10 ppm 2. 1 kg 1 L

concentration solvent solid concentrated solution quantitative solubility dilute pure

2

What Is It

The concentration of solution refers to the amount of solute present in a given

amount of solvent or solution. A solution can be qualitatively described as

• dilute: a solution that contains a small proportion of solute relative to solvent, or

• concentrated: a solution that contains a large proportion of solute relative to the solvent.

Quantitatively, one type of solution may be prepared and expressed in different

concentrations. One teaspoon of sugar in a cup of water is a different solution from

a cup of water with five teaspoons of sugar. The amount of solute in a solution may

be expressed in several ways. These include percent concentration - by mass, by

volume or by mass-volume, mole fraction, molality, molarity, and part per million.

A. Percent Concentration (by Mass, by Volume and by Mass-Volume)

I. Percent by Mass. This expresses the mass of solute per 100g of solution. In most applications, “percent concentration” means weight/weight percent (% weight/weight) which is equal to the number of grams of solute per 100 grams of solution. A solution that contains 30% by mass sugar means that the solution contains 30g of sugar dissolved in 70g of water. It also means that there are 30g of sugar per 100g of solution. The formula for percent by mass is:

% (𝐰𝐭/𝐰𝐭) = 𝐦𝐚𝐬𝐬 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐞

𝐦𝐚𝐬𝐬 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐞 + 𝐦𝐚𝐬𝐬 𝐨𝐟 𝐬𝐨𝐥𝐯𝐞𝐧𝐭 𝐱 𝟏𝟎𝟎

𝐦𝐚𝐬𝐬 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧 = 𝐦𝐚𝐬𝐬 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐞 + 𝐦𝐚𝐬𝐬 𝐨𝐟 𝐬𝐨𝐥𝐯𝐞𝐧𝐭 Sample Problem 1. If 7.5 g of sodium nitrate is dissolved in 85 ml of water, calculate the

concentration of sodium nitrate in the solution.

Solution: To find the total mass of solution, we must add the mass of solute, 7.5 grams, and the mass of water. Since the density of water is 1.0 g/ml, 85 ml of water is equivalent to 85 grams of water. Therefore, the total mass of the solution is 92.5 grams

% (𝐰𝐭/𝐰𝐭) = 𝐦𝐚𝐬𝐬 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐞

𝐦𝐚𝐬𝐬 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐞 + 𝐦𝐚𝐬𝐬 𝐨𝐟 𝐬𝐨𝐥𝐯𝐞𝐧𝐭 𝐱 𝟏𝟎𝟎

% (wt/wt) = 7.5 g

7.5 g + 85 g x 100

% (wt/wt) = 7.5 g

92.5 g x 100

% (wt/wt) = 0.08108108 x 100 % (wt/wt) = 8.1%

Lesson 1: CONCENTRATION OF SOLUTION

3

2. A common laboratory reagent is 10% (wt/wt) NaOH solution. How would you prepare 750 grams of the solution?

Solution: You can prepare 750 g of 10% (wt/wt) NaOH solutions as follows: a) You can translate the label 10% (wt/wt) NaOH as

𝟏𝟎𝐠 𝐍𝐚𝐎𝐇

𝟏𝟎𝟎 𝐠 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧 𝐨𝐫

𝟏𝟎𝟎 𝐠 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧

𝟏𝟎𝐠 𝐍𝐚𝐎𝐇

b) You can calculate the needed mass of NaOH using the first conversion factor

𝐠 𝐍𝐚𝐎𝐇 = 𝟕𝟓𝟎 𝐠 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧 𝐗 𝟏𝟎 𝐠 𝐍𝐚𝐎𝐇

𝟏𝟎𝟎 𝐠 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧

𝐠 𝐍𝐚𝐎𝐇 = 𝟕𝟓 𝐠 or you may solve it this way:

𝟏𝟎 % = 𝐠 𝐍𝐚𝐎𝐇

𝟕𝟓𝟎 𝐠 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧

𝐠 𝐍𝐚𝐎𝐇 = 𝟏𝟎%

𝟏𝟎𝟎% 𝐗 𝟕𝟓𝟎 𝐠

g NaOH = 0.10 X 750 g

𝐠 𝐍𝐚𝐎𝐇 = 𝟕𝟓 𝐠

Activity 3: Knowledge Check on Percent by Mass Concentration! Direction: Solve the given problem. You may use a separate sheet of paper.

II. Percent by Volume, Volume/Volume Percent or % (vol/vol)

When both solute and solvent are liquids, it is sometimes convenient

for you to describe the concentration as percent by Volume (%vol/vol) which is the number of Volume of the solute in 100 volume of solution. “Volume” may be any volume unit provided you use the same unit for both solute and solution.

% (𝐕𝐨𝐥/𝐕𝐨𝐥) = 𝐯𝐨𝐥𝐮𝐦𝐞 𝐨𝐟𝐬𝐨𝐥𝐮𝐭𝐞

𝐯𝐨𝐥𝐮𝐦𝐞 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧 𝐱 𝟏𝟎𝟎

𝐯𝐨𝐥𝐮𝐦𝐞 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧 = 𝐯𝐨𝐥𝐮𝐦𝐞 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐞 + 𝐯𝐨𝐥𝐮𝐦𝐞 𝐨𝐟 𝐬𝐨𝐥𝐯𝐞𝐧𝐭

If 7.5 g of sodium nitrate, NaNO3 is dissolved in 85ml of H2O (density of H2O = 1.0g/ml), calculate the %(wt/wt) concentration of H2O in the solution. Given:

Solution:

4

Sample Problem

1. A 40% (vol/vol) solution of ethylene glycol in water is used to give protection to a car’s cooling system. What Volume of ethylene glycol would you use to make five liters of this solution?

Solution: a) You can translate the label 40% (vol/vol) ethylene glycol as:

𝟒𝟎 𝐯𝐨𝐥 𝐞𝐡𝐭𝐲𝐥𝐞𝐧𝐞 𝐠𝐥𝐲𝐜𝐨𝐥

𝟏𝟎𝟎 𝐯𝐨𝐥 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧 𝐨𝐫

𝟏𝟎𝟎 𝐯𝐨𝐥 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧

𝟒𝟎 𝐯𝐨𝐥 𝐞𝐡𝐭𝐲𝐥𝐞𝐧𝐞 𝐠𝐥𝐲𝐜𝐨𝐥

In these factors, “vol” may be any volume unit you want – mL, liter,

or whatever is required. The problem specified liters, therefore you can use the second factor and compare the required volume as:

b) You can calculate the needed Volume of ethylene glycol using the first conversion factor

𝐥𝐢𝐭𝐞𝐫𝐬 𝐨𝐟 𝐞𝐭𝐡𝐲𝐥𝐞𝐧𝐞 𝐠𝐥𝐲𝐜𝐨𝐥 = 𝟓 𝐋 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧 𝐗 𝟒𝟎 𝐋 𝐞𝐭𝐡𝐲𝐥𝐞𝐧𝐞 𝐠𝐥𝐲𝐜𝐨𝐥

𝟏𝟎𝟎 𝐋 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧

𝐥𝐢𝐭𝐞𝐫𝐬 𝐨𝐟 𝐞𝐭𝐡𝐲𝐥𝐞𝐧𝐞 𝐠𝐥𝐲𝐜𝐨𝐥 = 𝟐 𝐋

or you may solve it this way:

𝟒𝟎 % = 𝐯𝐨𝐥𝐮𝐦𝐞 𝐨𝐟 𝐞𝐭𝐡𝐲𝐥𝐞𝐧𝐞 𝐠𝐥𝐲𝐜𝐨𝐥

𝟓 𝐋 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧

𝐯𝐨𝐥𝐮𝐦𝐞 𝐨𝐟 𝐞𝐭𝐡𝐲𝐥𝐞𝐧𝐞 𝐠𝐥𝐲𝐜𝐨𝐥 = 𝟒𝟎%

𝟏𝟎𝟎% 𝐗 𝟓 𝐋

𝐯𝐨𝐥𝐮𝐦𝐞 𝐨𝐟 𝐞𝐭𝐡𝐲𝐥𝐞𝐧𝐞 𝐠𝐥𝐲𝐜𝐨𝐥 = 0.40 X 5 L

𝐯𝐨𝐥𝐮𝐦𝐞 𝐨𝐟 𝐞𝐭𝐡𝐲𝐥𝐞𝐧𝐞 𝐠𝐥𝐲𝐜𝐨𝐥 = 𝟐 𝐋

Activity 4: Knowledge Check on Percent by Volume Concentration! Direction: Solve the given problem. You may use a separate sheet of paper.

III. Weight/Volume Percent or % (wt/vol)

When it is impractical to express both the solute and solvent in mass or volume units, a hybrid expression for percent is to be used. Hybrid because their units do not cancel as they ought to.

A brand of rubbing alcohol says, it contains 70% (vol/vol) isopropyl alcohol. How many mL of isopropyl alcohol are there in 600 ml of the solution in the bottle? Given:

Solution:

5

𝐏𝐞𝐫𝐜𝐞𝐧𝐭 𝐛𝐲 𝐦𝐚𝐬𝐬/𝐯𝐨𝐥𝐮𝐦𝐞 = 𝐦𝐚𝐬𝐬 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐞

𝐯𝐨𝐥𝐮𝐦𝐞 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧 𝐱 𝟏𝟎𝟎

Sample Problem

1. A solution is prepared by dissolving 5.0 grams of glucose in enough water to make 250 mL of solution. Calculate % (wt/vol) glucose.

Solution: Fill in the given to the equation,

% (𝐰𝐭/𝐯𝐨𝐥) = 𝐦𝐚𝐬𝐬 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐞

𝐯𝐨𝐥𝐮𝐦𝐞 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧 𝐱 𝟏𝟎𝟎

= 𝟓. 𝟎 𝐠

𝟐𝟓𝟎 𝐦𝐋 𝐱 𝟏𝟎𝟎

= 𝟐% 2. A 50 mL of 12% by mass-volume solution was used in an experiment. How

many grams of solute does the solution contain?

Solution: a) You can translate the label 12% (wt/vol) solution as

𝟏𝟐 𝐠 𝐬𝐨𝐥𝐮𝐭𝐞

𝟏𝟎𝟎 𝐦𝐋 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧 𝐨𝐫

𝟏𝟎𝟎 𝐦𝐋 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧

𝟏𝟐𝐠 𝐍𝐚𝐎𝐇

b) You can calculate the needed mass of the solute using the

first conversion factor

𝐠 𝐬𝐨𝐥𝐮𝐭𝐞 = 𝟓𝟎 𝐦𝐋 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧 𝐗 𝟏𝟐 𝐠 𝐬𝐨𝐥𝐮𝐭𝐞

𝟏𝟎𝟎 𝐦𝐋 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧

𝐠 𝐬𝐨𝐥𝐮𝐭𝐞 = 𝟔. 𝟎 𝐠

Activity 5: Knowledge Check on Percent by Mass/Volume Concentration! Direction: Solve the given problem. You may use a separate sheet of paper.

B. Mole Fraction, X

The variation in some physical and/or chemical properties of a solution especially those containing only two components is sometimes described over the entire range of concentration. The concentration of the solution is best described by mole fraction or mole percent.

Mole fraction is usually designated as X that relates to the number of

moles of a particular solute to the total number of moles in the solution.

The label of betadine skin cleanser says 7.5 % solution. Taking it to be % (wt/vol), how many grams of betadine (Providone-Iodine) are present in 50 mL bottle? Given:

Solution:

6

𝐌𝐨𝐥𝐞 𝐅𝐫𝐚𝐜𝐭𝐢𝐨𝐧 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐞, 𝐗(𝐬𝐨𝐥𝐮𝐭𝐞) = 𝐦𝐨𝐥𝐞 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐞

𝐦𝐨𝐥𝐞𝐬 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐞 + 𝐦𝐨𝐥𝐞𝐬 𝐨𝐟 𝐬𝐨𝐥𝐯𝐞𝐧𝐭

𝐌𝐨𝐥𝐞 𝐅𝐫𝐚𝐜𝐭𝐢𝐨𝐧 𝐨𝐟 𝐬𝐨𝐥𝐯𝐞𝐧𝐭, 𝐗(𝐬𝐨𝐥𝐯𝐞𝐧𝐭) = 𝐦𝐨𝐥𝐞 𝐨𝐟 𝐬𝐨𝐥𝐯𝐞𝐧𝐭

𝐦𝐨𝐥𝐞𝐬 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐞 + 𝐦𝐨𝐥𝐞𝐬 𝐨𝐟 𝐬𝐨𝐥𝐯𝐞𝐧𝐭

Sample Problem

1. Compute the mole fraction of acetone X (acetone) and of chloroform (X chloroform) in a solution prepared by mixing 50.0 g each of acetone (Molar Mass = 58.0) and chloroform (Molar Mass = 119.5)

Solution: a. The first step is for you to compute the number of moles (n) of

each substance as well as the total number of moles. (solute: acetone; solvent: chloroform)

𝐧 𝐨𝐟 𝐚𝐜𝐞𝐭𝐨𝐧𝐞 = 𝐦𝐚𝐬𝐬 𝐢𝐧 𝐠 𝐨𝐟 𝐚𝐜𝐞𝐭𝐨𝐧𝐞

𝐌𝐌 𝐨𝐟 𝐚𝐜𝐞𝐭𝐨𝐧𝐞

= 𝟓𝟎. 𝟎 𝐠 𝐚𝐜𝐞𝐭𝐨𝐧𝐞

𝟓𝟖 𝐠/𝐦𝐨𝐥 𝐚𝐜𝐞𝐭𝐨𝐧𝐞

= 𝟎. 𝟖𝟔𝟐 𝐦𝐨𝐥𝐞 𝐨𝐟 𝐚𝐜𝐞𝐭𝐨𝐧𝐞

𝐧 𝐨𝐟 𝐜𝐡𝐥𝐨𝐫𝐨𝐟𝐨𝐫𝐦 = 𝟓𝟎 𝐠 𝐨𝐟 𝐜𝐡𝐥𝐨𝐫𝐨𝐟𝐨𝐫𝐦

𝟏𝟏𝟗. 𝟓 𝐠/𝐦𝐨𝐥 𝐨𝐟 𝐜𝐡𝐥𝐨𝐫𝐨𝐟𝐨𝐫𝐦

= 𝟓𝟎. 𝟎 𝐠 𝐜𝐡𝐥𝐨𝐫𝐨𝐟𝐨𝐫𝐦

𝟏𝟏𝟗. 𝟓 𝐠/𝐦𝐨𝐥 𝐜𝐡𝐥𝐨𝐫𝐨𝐟𝐨𝐫𝐦

= 𝟎. 𝟖𝟔𝟐 𝐦𝐨𝐥𝐞 𝐨𝐟 𝐚𝐜𝐞𝐭𝐨𝐧𝐞

Total n in the solution = n of acetone + n of chloroform = 0.862 + 0.862

= 1.28 moles

b. Compute the X of acetone and X of chloroform

𝑿(𝐚𝐜𝐞𝐭𝐨𝐧𝐞) = 𝐧 𝐨𝐟 𝐚𝐜𝐞𝐭𝐨𝐧𝐞

𝐭𝐨𝐭𝐚𝐥 𝐧 𝐢𝐧 𝐭𝐡𝐞 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧

= 𝟎. 𝟎𝟖𝟐 𝐦𝐨𝐥

𝟏. 𝟐𝟖 𝐦𝐨𝐥

= 𝟎. 𝟔𝟕𝟑

𝑿(𝐜𝐡𝐥𝐨𝐫𝐨𝐟𝐨𝐫𝐦) = 𝐧 𝐨𝐟 𝐜𝐡𝐥𝐨𝐫𝐨𝐟𝐨𝐫𝐦

𝐭𝐨𝐭𝐚𝐥 𝐧 𝐢𝐧 𝐭𝐡𝐞 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧

= 𝟎. 𝟎𝟖𝟐 𝐦𝐨𝐥

𝟏. 𝟐𝟖 𝐦𝐨𝐥

= 𝟎. 𝟖𝟔𝟐

7

Activity 6: Knowledge Check on Mole Fraction! Direction: Solve the given problem. You may use a separate sheet of paper.

C. Molar Concentration or Molarity, M

Molarity is the ratio of the moles of solute to the volume of solution in

liters.

𝐌𝐨𝐥𝐚𝐫𝐢𝐭𝐲, 𝐌𝐨𝐥𝐚𝐫 𝐂𝐨𝐧𝐜𝐞𝐧𝐭𝐫𝐚𝐭𝐢𝐨𝐧, 𝐌 = 𝐦𝐨𝐥𝐞𝐬 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐞

𝐋𝐢𝐭𝐞𝐫 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧

𝐌 = 𝐧 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐞

𝐋 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧

Sample Problem

1. Calculate the molar concentration of the solution that contains 15 grams of potassium hydroxide (KOH) in 225 ml of solution. (Molar mass of KOH = 56 g/mol)

Solution: 1. Convert 15 grams of KOH to moles using conversion factor, 1

mol KOH = 56 g (the molar mass of KOH).

𝐦𝐨𝐥𝐞𝐬 𝐨𝐟 𝐊𝐎𝐇 = 𝟏𝟓 𝐠 𝐗𝟏 𝐦𝐨𝐥𝐞

𝟓𝟔 𝐠

= 0.268 moles

2. Convert 225 ml of solution to Liter of solution using the conversion factor: 1L = 1000 ml.

𝐋 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧 = 𝟐𝟐𝟓 𝐦𝐋 𝐗𝟏 𝐋

𝟏𝟎𝟎𝟎 𝐦𝐋

= 0.225 L Use the formula in computing molarity and substitute the values obtained above.

𝐌 = 𝐦𝐨𝐥𝐞𝐬 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐞

𝐋𝐢𝐭𝐞𝐫 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧

= 𝟎. 𝟐𝟔𝟖 𝐦𝐨𝐥

𝟎. 𝟐𝟐𝟓 𝐋

= 1.19 mol/L or 1.19 molar of 1.19 M

Calculate the mole fraction of sulfuric acid (H2SO4) in 8% (wt/wt) aqueous H2SO4 solution (molar masses: H2SO4 = 98 g/mol, H2O = 18 g/mol Given:

Solution:

8

Activity 7: Knowledge Check on Molarity! Direction: Solve the given problem. You may use a separate sheet of paper.

D. Molal Concentration or Molality, m

Solutions may also be expressed in molality. The molality of a

solution is the ratio of moles of solute to the mass of solvent in kilograms.

𝐦𝐨𝐥𝐚𝐥𝐢𝐭𝐲, 𝐦 = 𝐦𝐨𝐥𝐞𝐬 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐞

𝐤𝐢𝐥𝐨𝐠𝐫𝐚𝐦 𝐨𝐟 𝐬𝐨𝐥𝐯𝐞𝐧𝐭

Sample Problem 1. Determine the molal concentration, m of a solution that contains 18

grams of NaOH in 100 ml of water. The molar mass of NaOH is 40 g/mole. Solution: a. Convert 18 grams of NaOH into moles using the molar mass of NaOH

𝐦𝐨𝐥𝐞𝐬 𝐨𝐟 𝐍𝐚𝐎𝐇 = 𝟏𝟖 𝐠 𝐱 𝟏 𝐦𝐨𝐥𝐞

𝟒𝟎 𝐠

= 0.45 moles

b. Convert 100 ml of water into grams using the density of water, 1.0g/ml. Then convert the grams to kilograms using the conversion factor, 1Kg = 1000 g.

𝐊𝐠 𝐨𝐟 𝐬𝐨𝐥𝐯𝐞𝐧𝐭 = 𝟏𝟎𝟎 𝐦𝐋 𝐱 𝟏. 𝟎 𝐠

𝐦𝐋

= 𝟏𝟎𝟎 𝐠 𝐱 𝟏.𝟎 𝐤𝐠

𝟏𝟎𝟎𝟎 𝐠

= 𝟎. 𝟏𝐤𝐠 c. Use the formula in computing molality and substitute the

values obtained above.

𝐦 = 𝐦𝐨𝐥𝐞𝐬 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐞

𝐤𝐠 𝐨𝐟 𝐬𝐨𝐥𝐯𝐞𝐧𝐭

= 𝟎. 𝟒𝟓 𝐦𝐨𝐥

𝟎. 𝟏𝐤𝐠

= 4.5 mol/kg or 4.5 molal or 4.5 m

Calculate the molar concentration of a solution that contains 23 g of potassium hydroxide. KOH in 250 ml of solution. Molar mass of KOH is 56 g/mol. Given:

Solution:

9

Activity 8: Knowledge Check on Molality! Direction: Solve the given problem. You may use a separate sheet of paper.

E. Parts per Million, ppm

For a very dilute solution where the concentration of the solute is very low

like pesticide residue in water or heavy metal like Hg2+ concentration in effluents, or even hardness of water, it is convenient to express concentration in terms of parts per million, ppm.

Parts per million (ppm) expresses the number of parts of solute per one

million parts of the solution.

𝐏𝐚𝐫𝐭𝐬 𝐩𝐞𝐫 𝐦𝐢𝐥𝐥𝐢𝐨𝐧 = 𝐦𝐚𝐬𝐬 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐞

𝐯𝐨𝐥𝐮𝐦𝐞 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧 𝐱 𝟏, 𝟎𝟎𝟎, 𝟎𝟎𝟎

𝐩𝐩𝐦 =𝐠𝐫𝐚𝐦𝐬 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐞

𝟏𝟎𝟎𝟎𝟎𝟎𝟎 𝐠 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧 =

𝐦𝐢𝐥𝐥𝐢𝐠𝐫𝐚𝐦 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐞

𝐤𝐢𝐥𝐨𝐠𝐫𝐚𝐦 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧=

𝐦𝐢𝐥𝐥𝐢𝐠𝐫𝐚𝐦 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐞

𝐥𝐢𝐭𝐞𝐫 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧

The last expression is approximately true for water as a solvent because the density of water is 1.0 g/ml. Also, for a very dilute solution, the amount of the solution could be equated to the amount of the solvent, water.

While these concentrations are very small, but we should not neglect their importance. Some of the industrial pollutants that are being released daily into the water we drink and the air we breathe can be extremely harmful in concentrations as small as 1 ppm.

Sample Problem 1. A water sample was reported to contain 250 ppm CaCO3. How many

grams of CaCO3 is present in 4 liters of water. Solution: 250 ppm CaCO3 can be translated as 250 mg/liter of solution. Since this is a very dilute solution and the solvent is water, the liter of solution could be equated to the Volume of water. So, we can use the expression below,

𝐩𝐩𝐦 =𝐦𝐢𝐥𝐥𝐢𝐠𝐫𝐚𝐦 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐞

𝐥𝐢𝐭𝐞𝐫 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧= 𝟐𝟓𝟎 𝐩𝐩𝐦 =

𝟐𝟓𝟎 𝐦𝐠

𝟏 𝐋

The problem asks for the mass in grams of CaCO3 present in 4 liters of

water:

𝐦𝐠 𝐨𝐟 𝐂𝐚𝐂𝐎𝟑 = 𝟒 𝐋 𝐗 𝟐𝟓𝟎 𝐦𝐠

𝟏 𝐋

= 𝟏𝟎𝟎𝟎 𝐦𝐠

How many grams of solute is present in 0.4 moles Magnesium hydroxide, Mg(OH)2 in 550 g water, H2O? The molar mass of Mg(OH)2 is 58 g/mol.

Given:

Solution:

10

Activity 9: Knowledge Check on Molality! Direction: Solve the given problem. You may use a separate sheet of paper.

Activity 10: Remember Me! Direction: Identify the concept related to solution stoichiometry that is described in each statement Get your answer from the word bank.

_______________1. It states that the mass of the products is equal to the mass of the

reactants.

_______________2. It states that the mass of one element combines with a fixed mass of

another element in a ratio of whole numbers

_______________3. It states that all samples of a given chemical compound have the

same elemental composition

_______________4. For a chemical equation to be correct, it must be __.

_______________5. A number written before the symbol of an element or formula of a

compound.

Activity 10 enabled you to recall important concepts and laws related to

chemical equations. Mass relations are based on the three important laws. If you keep

these laws in mind, you'll be able to make valid predictions and calculations for

chemical reactions including those that involve solutions.

Stoichiometry Involving Solution

In General Chemistry 1, you have done several calculations using a balanced

chemical equation. Stoichiometry deals with solving quantitative problems using a balanced chemical equation. Recall that in these types of calculations, we used the following steps:

1. Convert from the given units to moles, if not given in moles; 2. Convert from moles of the given quantity to moles of the desired quantity,

using the balanced equation; and 3. Convert from moles to any other desired units.

Law of Multiple Proportion superscript Law of Constant Composition balanced Hess Law coefficient Law of Definite Proportion subscript Law of Conservation of Mass

A commercial pesticide formulation contains 1.0 g deltametrin in 1L solution. What is its concentration in ppm? Given:

Solution:

Lesson 2: SOLUTION STOICHIOMETRY

11

The number of moles of a substance can be related to its molar mass and number of molecules. It can also be related to the volume at Standard Temperature and Pressure (STP). Sample Problem

1. What volume of 0.556 M HCl has enough HCl to combine exactly with 25.4 mL of 0.458 M NaOH? The equation for the reaction is,

HCl(aq) + NaOH(aq) → NaCl (aq) + H2O (l)

Solution: 1. Find the moles of NaOH in 25.4 mL of 0.458 M NaOH 2. From the expression of molarity, M = n of solute / L of solution 3. The molarity of the solution gives you two conversion factors:

𝟎. 𝟒𝟓𝟖 𝐦𝐨𝐥𝐞𝐬 𝐍𝐚𝐎𝐇

𝟏𝟎𝟎𝟎 𝐦𝐋 𝐍𝐚𝐎𝐇 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧 𝐨𝐫

𝟏𝟎𝟎𝟎 𝐦𝐋 𝐍𝐚𝐎𝐇 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧

𝟎. 𝟒𝟓𝟖 𝐦𝐨𝐥𝐞𝐬 𝐍𝐚𝐎𝐇

4. When you use the first conversion factor, you can get the moles of NaOH as:

𝐦𝐨𝐥𝐞𝐬 𝐍𝐚𝐎𝐇 = 𝟐𝟓. 𝟒 𝐦𝐋 𝐍𝐚𝐎𝐇 𝐗 𝟎. 𝟒𝟓𝟖 𝐦𝐨𝐥 𝐍𝐚𝐎𝐇

𝟏𝟎𝟎𝟎 𝐦𝐋 𝐍𝐚𝐎𝐇 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧

= 0.0116 moles

5. Use the coefficients of NaOH and HCl from the balanced equation to calculate how much 0.0116 mole NaOH is equivalent to in mole of HCl.

From the balanced equation, the ratio of coefficients of NaOH and HCl is 1:1

𝐦𝐨𝐥𝐞𝐬 𝐨𝐟 𝐇𝐂𝐥 = 𝟎. 𝟎𝟏𝟏𝟔 𝒎𝒐𝒍𝒆 𝐍𝐚𝐎𝐇 𝐗 𝟏 𝐦𝐨𝐥 𝐇𝐂𝐥

𝟏𝐦𝐨𝐥 𝐍𝐚𝐎𝐇

= 𝟎. 𝟎𝟏𝟏𝟔 𝒎𝒐𝒍𝒆 HCl

6. Find the Volume of HCl using 0.556 M, the given molarity of HCl aqueous solution.

𝐯𝐨𝐥𝐮𝐦𝐞 𝐨𝐟 𝐇𝐂𝐥 = 𝟎. 𝟎𝟏𝟏𝟔 𝒎𝒐𝒍𝒆 𝐇𝐂𝐥 𝐗 𝟏𝟎𝟎𝟎 𝐦𝐋 𝐇𝐂𝐥

𝟎. 𝟓𝟓𝟔 𝐦𝐨𝐥 𝐇𝐂𝐥

= 𝟐𝟎. 𝟗 𝐦𝐋 𝐇𝐂𝐥

2. Calculate the mass (in grams) of calcium nitrate, Ca(NO3)2 that can be produced by reacting 136 ml of 4.00 M nitric acid, HNO3 with excess calcium hydroxide, Ca(OH)2. The molar mass of Ca(NO3)2 = 164 g/mol

Solution:

1. Write the balanced equation for the reaction

2HNO3 + Ca(OH)2 → 2H2O + Ca(NO3)2

2. Use the molarity and Volume of the solution to get the number of moles of HNO3.

Moles of HNO3 = (molarity) (Volume of the solution in liters) = (4.0 mole/L) (0.136 L) = 0.544 mol

12

3. Find the number of moles of Ca(NO3)2 using the stoichiometric factor:

𝐦𝐨𝐥𝐞𝐬 𝐨𝐟 𝐂𝐚(𝐍𝐎𝟑)𝟐 = 𝟎. 𝟓𝟒𝟒 𝒎𝒐𝒍𝒆 𝐇𝐍𝐎𝟑 𝐗 𝟏 𝐦𝐨𝐥 𝐂𝐚(𝐍𝐎𝟑)𝟐

𝟐𝐦𝐨𝐥 𝐇𝐍𝐎𝟑

4. Find the mass of Ca(NO3)2 using its molar mass:

𝐦𝐚𝐬𝐬 𝐨𝐟 𝐂𝐚(𝐍𝐎𝟑)𝟐 = 𝟎. 𝟐𝟕𝟐 𝒎𝒐𝒍𝒆𝐂𝐚(𝐍𝐎𝟑)𝟐 𝐗 𝟏𝟔𝟒 𝐠 𝐂𝐚(𝐍𝐎𝟑)𝟐

𝟏𝐦𝐨𝐥 𝐂𝐚(𝐍𝐎𝟑)𝟐

= 𝟒𝟒. 𝟔 𝒈 Ca(NO3)2

Activity 11: Knowledge Check on Solution Stoichiometry!

Direction: Solve the given problem. You may use a separate sheet of paper.

Colligative properties of solutions are properties that depend upon the concentration of solute molecules or ions, but not upon the identity of the solute. Colligative properties include vapor pressure lowering, boiling point elevation, freezing point depression, and osmotic pressure.

Lowering the Vapor Pressure

Vapor pressure is the pressure of a vapor in thermodynamic equilibrium with its condensed phase in a closed container. When a non-volatile solute is dissolved in a solvent, the vapor pressure of the solvent is lowered. The presence of solute decreases the rate of escape of solvent molecules resulting in lower vapor pressure.

Boiling Point Elevation

The boiling point of a liquid is defined as the temperature at which the vapor pressure of that liquid equals the atmospheric pressure (760mm Hg). The addition of the solute increases the boiling point of the solution. The atmospheric pressure remains the same while the vapor pressure of the solution is lowered resulting in the increase of the difference in atmospheric pressure and vapor pressure of the solution. Therefore, a higher temperature is required to boil the solution.

Freezing Point Depression

Normal freezing or melting point is the temperature at which solid and liquid are in equilibrium under 1 atm. The addition of solute will decrease the vapor pressure and so will decrease the freezing point.

For a liquid to freeze it must achieve a very ordered state that results in the formation of a crystal. If there are impurities in the liquid, i.e. solutes, the liquid is inherently less ordered. The presence of impurities in a liquid or a substance makes a

variation in the freezing point by making them low or high. Therefore, a solution is more difficult to freeze than a pure solvent so a lower temperature is required

to freeze the liquid.

How many mL of 0.250 M HCl would react exactly with 30.0 mL of the 0.150 M solution of Ca(OH)2 solution? The chemical reaction involved is:

HCl(aq) + NaOH(aq) → NaCl (aq) + H2O (l) Given:

Solution:

Lesson 3: EFFECTS OF CONCENTRATION ON THE COLLIGATIVE PROPERTIES OF SOLUTION

13

Osmotic Pressure This is the external pressure that must be applied to the solution to prevent it

from being diluted by the entry of solvent via osmosis. Osmosis is the movement of solvent particles across a semipermeable membrane from a dilute solution (low concentration) into a concentrated solution. The solvent moves to dilute the concentrated solution and equalize the concentration on both sides of the membrane.

Osmotic pressure is directly proportional to the concentration of the solution.

Therefore, doubling the concentration will also double the osmotic pressure. The osmotic pressure of two solutions having the same molal concentration is identical. Activity 12: True or False Direction: Read each statement and evaluate if it is correct (True) or not (False). Write your answer in the given space.

_______1. Colligative properties arise from the fact that solute affects the concentration of solvent.

_______2. Vapor pressure is a colligative property. _______3. Lowering of vapor pressure is not dependent on the number of species present in the solution. _______4. The colligative properties of solutions depend on the nature of the solute and the solvent. _______5. Colligative molality is the molality times the number of solute particles per formula unit. _______6. Osmotic pressure is directly proportional to the concentration of the solution. ______ 7. Relative lowering of vapor pressure is a colligative property. ______ 8. The boiling point of a solution decreases in direct proportion to the molality of the solute. ______ 9. When a non-volatile solute is dissolved in a solvent, the vapor pressure of the solvent is lowered.

______10.The depression of the freezing point is directly proportional to the molality of the solvent.

What’s More

Let’s talk about preparing a solution. Direction: Use a separate sheet of paper for your answers to each of these problems. 1. A common laboratory reagent is 10% (wt/wt) NaOH solution. Describe how would

you prepare 750 g of this solution?

2. How would you prepare 250 mL of 0.200 M sodium hydroxide, NaOH solution?

Show the details of your computations and describe exactly how you will prepare the solution.

3. Comment on the way the solution is prepared. To make 1.00 L of a 0.05 molal sucrose, C12H22O11 solution you need to dissolve 25 grams of sucrose. Do a detailed computation to describe how the solution should be prepared. You may also use an illustration.

14

What I Have Learned

Solve the following problems and show your solution.

1. What mass of NaCl is needed to make a 1.5 m solution using 300 g of water as a solvent?

2. A bottle of wine label reads 12% alcohol by volume. Calculate the volume in mL of alcohol in 350 ml of wine.

3. How many grams of H3PO4 are in 175 mL of a 3.5 M solution of H3PO4 4. Ammonia and phosphoric acid solutions are used to produce ammonium

hydrogen phosphate fertilizer. What Volume of 14.8 M NH3(aq) is needed to

react with 100 L of 12.9 M of H3PO4(aq)?

What I Can Do Describe the effects of concentration on the colligative property of the solution. The 1st answer identifies the colligative property involved while the 2nd answer provides the reason on how it affects the property of the solution.

QUESTION COLLIGATIVE PROPERTY AFFECTED AND

HOW IT IS AFFECTED

Adding salt to water increases the boiling point?

1._______________________________________ 2._______________________________________ _________________________________________

Ice water gets colder when salt is added?

3._______________________________________ 4._______________________________________ _________________________________________

What colligative property is dominantly effected when urea

is added to water.

5._______________________________________ 6._______________________________________

_________________________________________

Why does seawater have a lower freezing point than pure water?

7._______________________________________ 8._______________________________________ ________________________________________

How can you regain the crispness of a carrot and celery that have become limp?

9._______________________________________ 10.______________________________________ _________________________________________

15

Assessment Direction: Encircle the letter of the best answer. 1. A solution contains 28% phosphoric acid by mass. This means that:

A. 100 ml of this solution contains 28 g of phosphoric acid B. 1 ml of this solution contains 28 g of phosphoric acid C. 1 L of this solution contains 28 mL of phosphoric acid D. 1 L of this solution has a mass of 28 g

2. Calculate the concentration in % (wt/wt) of a solution containing 20.0 g of NaCl

dissolved in 250.0 g of H2O. A. 6.76% (m/m) C. 8.00% (m/m) B. 7.41% (m/m) D. 8.25% (m/m)

3. What is the concentration in % (m/v) of a NaCl solution prepared by dissolving 9.3 g of NaCl in a sufficient amount of water to give 350 mL of solution?

A. 3.26% (m/v) C. 37.6% (m/v) B. 0.455% (m/v) D. 2.66 (m/v)

4. Calculate the grams of NaOH present in 5.0 mL of a 1.0% (m/v) NaOH solution. A. 0.050 g C. 0.50 g B. 0.10 g D. 1.0 g

5. How many grams of NaOH are there in 500.0 mL of a 0.175 M NaOH solution? A. 14 g C. 114 g B. 3.50 g D. 0.00219 g

6. What is the molarity of an aqueous solution containing 22.5 g of sucrose (C12H22O11)

in 35.5 mL of solution? A. 1.85 M C. 1.85 M B. 1.85 m D. 1.85 m

7. How many grams of H3PO4 are in 175 mL of a 3.5 M solution of H3PO4? A. 4.9 C. 60 B. 20 D. 612

8. What is the molality of 6 grams of table salt, NaCl in 10 grams of a solution? (MM of NaCl = 58.45 g/mol)

A. 1.027 m C. 1.027 m B. 10.27 M D. 10.27 M

9. What is the molality of an aqueous NaOH solution made with dissolving 5.0

Kg of water and 3.6 moles of NaOH? A. 3.6 m C. 1.4 m B. 0.72 m D. 0.090 m

10. After mixing 10.00 g of compound A with 20.00 g of compound B, it is found that

the mole fraction of compound A is 0.400. The mole fraction of compound B must be:

A. 0.200 C. 0.600 B. 0.400 D. 0.800

11. What is the mole fraction of CaCl2 (molar mass=111 g/mole) when 3.75 g of it is

placed in 10.1 g of water (molar mass=18.0 g/mole)? A. 0.8752 C. 0.1043 B. 0.5280 D. 0.0568

16

12. How many grams of Calcium phosphate, Ca3(PO4)2 can be produced from the reaction of 2.50 L of 0.250 M Calcium chloride, CaCl2 with an excess Phosphoric acid, H3PO4. The balanced equation is,

3CaCl2 + 2H3PO4 → Ca3(PO4)2 + 6HCl A. 34.4 g C. 76.4 g B. 46.6 g D. 64.6 g

13. How many liters of 0.53 M HCl is required to neutralize 0.78 g sodium

carbonate, Na2CO3? The balanced equation is 2HCl + NaCO3 → 2NaCl + H2CO3

A. 0.028 L C. 1.128 L B. 0.082 L D. 1.182 L

14. Adding salt to water will make the freezing point of the resulting solution

______0oC.

A. equal to C. lower than B. higher than D. greater than or equal to

15. As the temperature increases, the solubility of a gas in a carbonated drink

_________. A. decreases C. stays the same B. increases D. decreases then increases

Additional Activity

Activity 6: Which ice cube will melt first? Source: How Substances Affect Water’s Freezing Point available at https://www.clearwaycommunitysolar.com/blog/ science-center-home-experiments-for-kids/global-warming-ice-melting-experiment-for-kids/

Before beginning, think about the following questions and write down your predictions:

1. Do you think all of the ice cubes will melt at the same time? Why? 2. If you said “no” to question 1, which ice block do you think will be the first to

completely finish melting? 3. Do you think there will be a big difference between the ice block that melts the

fastest and the ice block that melts the slowest?

Materials Needed:

3 oz cups (can be plastic or paper) Water 1 tablespoon of the following: Marking pen

Table Salt Freezer (0°C) Granulated Sugar Paper and pen Flour Stopwatch

Procedure: 1. Measure one tablespoon of each substance and place in four separate cups. 2. Label each cup with the name of the substance it contains. 3. Pour an equal amount of water into each cup. Fill the cups so that they are almost full to the top. Stir very gently until the substances are fully dissolved (note: the flour and water mixture will turn into pancake batter). Be careful not to spill any water out of the cups while stirring. 4. Place the cups in a freezer for 2 hours. Make sure the freezer is set to 0°C or colder. 5. While you wait for the ice cubes to form, create a chart where you can record the time (minutes) it takes for each ice cube sample to melt.

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6. Remove each cup from the freezer and place it outside in the sun. 7. Press “start” on your stopwatch and watch as your ice samples start to melt. When one of the ice cubes melts completely, record the time under the correct column for that sample. Keep the timer running and record the time it takes for each ice cube to fully melt. You may stop the stopwatch after the last ice cube melts. Share Your Results

1. Which ice cube sample melted the fastest? Why?

2. Did the flour cause the ice to melt faster? Why not? (use Google to help find the answer if you need to).

Answer Key General Chemistry 2 Module 3

WATER SALT FLOUR SUGAR

______ min. ______ min. _____ min. ______ min.

What’s In: Activity 1 1. concentration 2. solvent 3. concentrated 4. solution 5. solubility 6. dilute 7. quantitative

What’s New: Activity 2

1. 0.350 mol 2. 10 mg

3. 80 ml 4. 1 Kg 5. 100 ml

What Is It Activity 3: 8.1% Activity 4: 3.50 ml Activity 5: 3.75 g Activity 6: 0.01 Activity 7: 1.79 M Activity 8: 12.8 g Activity 9: 1000 ppm Activity 10

1. Law of conservation of mass

2. Law of multiple proportion

3. Law of constant composition

4. Balanced 5. Coefficient

Activity 11: 36.0 ml HCl Activity 12

1. True

2. False 3. False 4. False 5. True 6. True 7. True 8. False 9. True 10. False

Assessment 1. A 2. B 3. D 4. A 5. B 6. A 7. C 8. A 9. B 10. C

11. D 12. D 13. A 14. C 15. A Additional Activity Salt lowers the freezing/melting point of water/ice. When the salty ice cube was placed in the sun, the lower freezing point combined with continuous heat from the sun made the ice melt much faster.

Sugar is also soluble in water, and also lowered the freezing/melting point of the water, but sugar does not make ice melt as fast as salt does. Flour does not cause the ice cube to melt faster because the flour has almost the

same freezing/melting point as pure water.

What’s More 1. To prepare 750g solution of

10% (wt/wt) NaOH solution you need to dissolve 75g of NaOH.

2. You should dissolve 4.2g of NaOH in enough water to make 250 ml solution

3. No. Because to prepare 0.05 molar sucrose solution, only 17.1g of sucrose is needed.

What I Have Learned 1. 88.125 g NaCl 2. 42 ml 3. 60.025 g 4. 1,740 L What I Can Do 1. boiling point elevation 2. The addition of these particles

raises the boiling point because part of the pressure that the solution exerts on the atmosphere come not only from water but from the solute particles

3. freezing point depression 4. dissolving salt in water lower’s

the liquids freezing point

5. vapor pressure 6. increasing urea concentration

decreases the evaporation rate and drop of temperature

7. freezing point depression 8. pressure of salt that

dissociates its component ion.

9. osmosis

10. by placing it in fresh

water the carrot and celery will absorb the water

restoring its crispness

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References

Ayson, M, et al. General Chemistry 2 Senior High School Textbook, 2016 Bayquen, A and Gardee Peña. General Chemistry 2. Phoenix Publishing House, 2016. Fajardo, N and Macario Catahan. Chem C8 Chemistry and the Environment, UPOU

Learning Resource, 1994 Hagad, Hilda R., Phoenix Next Century Chemistry, 2003 Mortimer Charles E., Chemistry 6th Edition Rodriquez, M, and Ma. Cecilia de Mesa. Fundamental Concepts of Chemistry II, UPOU

Learning Resource, 1994 Santos, Gil Nonato S., Danac, Alfonso C., O-Chemistry III, 2009

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