what volume of 10.2m hcl is required to prepare 50 ml of a 0.500 m hcl solution?

Affiliate 3. Composition of Substances and Solutions

three.iii Molarity

Learning Objectives

By the terminate of this section, you will be able to:

• Draw the primal properties of solutions
• Calculate solution concentrations using molarity
• Perform dilution calculations using the dilution equation

In preceding sections, we focused on the composition of substances: samples of matter that contain only i type of chemical element or compound. However, mixtures—samples of matter containing two or more substances physically combined—are more commonly encountered in nature than are pure substances. Similar to a pure substance, the relative composition of a mixture plays an important role in determining its properties. The relative amount of oxygen in a planet's atmosphere determines its ability to sustain aerobic life. The relative amounts of iron, carbon, nickel, and other elements in steel (a mixture known equally an "blend") determine its physical strength and resistance to corrosion. The relative amount of the active ingredient in a medicine determines its effectiveness in achieving the desired pharmacological effect. The relative amount of sugar in a beverage determines its sweetness (see Figure 1). In this section, we will describe one of the most common ways in which the relative compositions of mixtures may be quantified.

Figure ane. Sugar is one of many components in the circuitous mixture known as coffee. The amount of sugar in a given amount of coffee is an important determinant of the drinkable's sweetness. (credit: Jane Whitney)

Solutions

We have previously divers solutions every bit homogeneous mixtures, meaning that the limerick of the mixture (and therefore its properties) is uniform throughout its entire volume. Solutions occur frequently in nature and have also been implemented in many forms of manmade engineering science. We will explore a more thorough treatment of solution properties in the affiliate on solutions and colloids, but here we will introduce some of the basic properties of solutions.

The relative corporeality of a given solution component is known as its concentration. Oft, though not always, a solution contains i component with a concentration that is significantly greater than that of all other components. This component is chosen the solvent and may be viewed as the medium in which the other components are dispersed, or dissolved. Solutions in which water is the solvent are, of course, very common on our planet. A solution in which h2o is the solvent is called an aqueous solution.

A solute is a component of a solution that is typically present at a much lower concentration than the solvent. Solute concentrations are often described with qualitative terms such as dilute (of relatively low concentration) and full-bodied (of relatively loftier concentration).

Concentrations may be quantitatively assessed using a wide multifariousness of measurement units, each user-friendly for particular applications. Molarity (K) is a useful concentration unit of measurement for many applications in chemistry. Molarity is defined as the number of moles of solute in exactly 1 liter (i L) of the solution:

[latex]M = \frac{\text{mol solute}}{\text{L solution}}[/latex]

Case ane

Calculating Molar Concentrations
A 355-mL soft drink sample contains 0.133 mol of sucrose (table sugar). What is the tooth concentration of sucrose in the potable?

Solution
Since the tooth amount of solute and the volume of solution are both given, the molarity can be calculated using the definition of molarity. Per this definition, the solution book must be converted from mL to L:

[latex]M = \frac{\text{mol solute}}{\text{L solution}} = \frac{0.133 \;\text{mol}}{355 \;\text{mL} \times \frac{ane \;\text{50}}{chiliad \;\text{mL}}} = 0.375 \; M[/latex]

Check Your Learning
A teaspoon of table carbohydrate contains most 0.01 mol sucrose. What is the molarity of sucrose if a teaspoon of carbohydrate has been dissolved in a cup of tea with a volume of 200 mL?

Example 2

Deriving Moles and Volumes from Molar Concentrations
How much sugar (mol) is contained in a modest sip (~x mL) of the soft drink from Example i?

Solution
In this case, nosotros tin rearrange the definition of molarity to isolate the quantity sought, moles of carbohydrate. We then substitute the value for molarity that we derived in Instance 1, 0.375 M:

[latex]M = \frac{\text{mol solute}}{\text{L solution}}[/latex]
[latex]\text{mol solute} = Yard \times \text{50 solution}[/latex]

[latex]\text{mol solute} = 0.375 \;\frac{\text{mol sugar}}{\text{50}} \times (ten \;\text{mL} \times \frac{1 \text{Fifty}}{1000 \;\text{mL}}) = 0.004 \;\text{mol sugar}[/latex]

Check Your Learning
What volume (mL) of the sweetened tea described in Example 1 contains the same amount of saccharide (mol) as 10 mL of the soft drinkable in this example?

Example 3

Calculating Molar Concentrations from the Mass of Solute
Distilled white vinegar (Figure two) is a solution of acetic acid, CH₃CO₂H, in water. A 0.500-L vinegar solution contains 25.2 g of acetic acid. What is the concentration of the acetic acid solution in units of molarity?

Figure 2. Distilled white vinegar is a solution of acerb acid in h2o.

Solution
As in previous textbox shaded, the definition of molarity is the master equation used to calculate the quantity sought. In this case, the mass of solute is provided instead of its molar amount, then we must utilize the solute's tooth mass to obtain the amount of solute in moles:

[latex]Thousand = \frac{\text{mol solute}}{\text{L solution}} = \frac{25.2 \;\text{thou CH}_3\text{CO}_2\text{H} \times \frac{one \;\text{mol CH}_2\text{CO}_2\text{H}}{60.052 \;\text{chiliad CH}_2\text{CO}_2\text{H}}}{0.500 \;\text{L solution}} = 0.839 \;Yard[/latex]

[latex]\begin{assortment}{r @{{}={}} l} M & \frac{\text{mol solute}}{\text{L solution}} = 0.839\;M \\[1em] M & \frac{0.839 \;\text{mol solute}}{one.00 \;\text{L solution}} \finish{assortment}[/latex]

Check Your Learning
Calculate the molarity of 6.52 yard of CoCl_ii (128.9 thousand/mol) dissolved in an aqueous solution with a full volume of 75.0 mL.

Example 4

Determining the Mass of Solute in a Given Book of Solution
How many grams of NaCl are independent in 0.250 L of a 5.30-G solution?

Solution
The volume and molarity of the solution are specified, so the corporeality (mol) of solute is hands computed as demonstrated in Example 2:

[latex]M = \;\frac{\text{mol solute}}{\text{L solution}}[/latex]
[latex]\text{mol solute} = Yard \times \text{50 solution}[/latex]
[latex]\text{mol solute} = 5.30 \;\frac{\text{mol NaCl}}{\text{L}} \times 0.250 \;\text{L} = one.325 \;\text{mol NaCl}[/latex]

Finally, this molar amount is used to derive the mass of NaCl:

[latex]ane.325 \;\text{mol NaCl} \times \frac{58.44 \;\text{g NaCl}}{\text{mol NaCl}} = 77.4 \;\text{g NaCl}[/latex]

Cheque Your Learning
How many grams of CaCl_two (110.98 grand/mol) are contained in 250.0 mL of a 0.200-Grand solution of calcium chloride?

When performing calculations stepwise, as in Example 4, it is important to refrain from rounding whatever intermediate calculation results, which tin pb to rounding errors in the final upshot. In Example 4, the molar amount of NaCl computed in the first step, 1.325 mol, would be properly rounded to ane.32 mol if information technology were to exist reported; however, although the last digit (5) is not significant, it must exist retained equally a guard digit in the intermediate calculation. If we had non retained this guard digit, the last calculation for the mass of NaCl would have been 77.1 g, a difference of 0.3 k.

In add-on to retaining a guard digit for intermediate calculations, we can too avoid rounding errors by performing computations in a single step (see Example 5). This eliminates intermediate steps so that just the final result is rounded.

Example five

Determining the Volume of Solution Containing a Given Mass of Solute
In Example 3, nosotros plant the typical concentration of vinegar to exist 0.839 1000. What volume of vinegar contains 75.six g of acetic acid?

Solution
Starting time, utilize the molar mass to calculate moles of acerb acid from the given mass:

[latex]\text{thousand solute} \times \frac{\text{mol solute}}{\text{g solute}} = \text{mol solute}[/latex]

And so, utilize the molarity of the solution to calculate the volume of solution containing this molar amount of solute:

[latex]\text{mol solute} \times \frac{\text{L solution}}{\text{mol solute}} = \text{Fifty solution}[/latex]

Combining these ii steps into i yields:

[latex]\text{g solute} \times \frac{\text{mol solute}}{\text{g solute}} \times \frac{\text{L solution}}{\text{mol solute}} = \text{L solution}[/latex][latex]75.6 \;\text{thou CH}_3\text{CO}_2\text{H} (\frac{\text{mol CH}_3\text{CO}_2\text{H}}{60.05 \;\text{1000}}) (\frac{\text{Fifty solution}}{0.839 \;\text{mol CH}_3\text{CO}_2\text{H}}) = one.fifty \;\text{Fifty solution}[/latex]

Check Your Learning
What volume of a 1.l-M KBr solution contains 66.0 chiliad KBr?

Dilution of Solutions

Dilution is the process whereby the concentration of a solution is lessened past the addition of solvent. For example, nosotros might say that a glass of iced tea becomes increasingly diluted as the ice melts. The water from the melting water ice increases the volume of the solvent (water) and the overall volume of the solution (iced tea), thereby reducing the relative concentrations of the solutes that requite the beverage its taste (Figure 3).

Figure 3. Both solutions contain the same mass of copper nitrate. The solution on the correct is more dilute considering the copper nitrate is dissolved in more solvent. (credit: Mark Ott)

Dilution is too a mutual means of preparing solutions of a desired concentration. Past adding solvent to a measured portion of a more full-bodied stock solution, we tin can achieve a particular concentration. For example, commercial pesticides are typically sold every bit solutions in which the active ingredients are far more full-bodied than is advisable for their application. Before they tin be used on crops, the pesticides must be diluted. This is also a very common practice for the training of a number of common laboratory reagents (Effigy 4).

Figure 4. A solution of KMnO₄ is prepared by mixing water with 4.74 one thousand of KMnO₄ in a flask. (credit: modification of piece of work by Mark Ott)

A simple mathematical human relationship can be used to relate the volumes and concentrations of a solution before and subsequently the dilution process. According to the definition of molarity, the tooth amount of solute in a solution is equal to the product of the solution's molarity and its volume in liters:

[latex]n = ML[/latex]

Expressions like these may exist written for a solution before and afterward it is diluted:

[latex]n_1 = M_1L_1[/latex]

[latex]n_2 = M_2L_2[/latex]

where the subscripts "one" and "2" refer to the solution before and after the dilution, respectively. Since the dilution process does not change the corporeality of solute in the solution, north ₁ = northward _two. Thus, these two equations may be set equal to one another:

[latex]M_1L_1 = M_2L_2[/latex]

This relation is commonly referred to every bit the dilution equation. Although we derived this equation using molarity equally the unit of concentration and liters every bit the unit of book, other units of concentration and volume may exist used, so long as the units properly cancel per the gene-characterization method. Reflecting this versatility, the dilution equation is often written in the more than general class:

[latex]C_1V_1 = C_2V_2[/latex]

where C and V are concentration and book, respectively.

Utilise the simulation to explore the relations between solute corporeality, solution volume, and concentration and to confirm the dilution equation.

Example 6

Determining the Concentration of a Diluted Solution
If 0.850 L of a five.00-1000 solution of copper nitrate, Cu(NO₃)_two, is diluted to a book of one.80 Fifty past the improver of water, what is the molarity of the diluted solution?

Solution
We are given the volume and concentration of a stock solution, Five _i and C ₁, and the volume of the resultant diluted solution, Five _two. We need to detect the concentration of the diluted solution, C ₂. Nosotros thus rearrange the dilution equation in order to isolate C ₂:

[latex]C_1V_1 = C_2V_2[/latex]
[latex]C_2 = \frac{C_1V_1}{V_2}[/latex]

Since the stock solution is being diluted by more than two-fold (volume is increased from 0.85 L to i.80 L), nosotros would await the diluted solution's concentration to be less than i-half five Yard. Nosotros volition compare this ballpark estimate to the calculated consequence to check for any gross errors in computation (for case, such every bit an improper substitution of the given quantities). Substituting the given values for the terms on the right side of this equation yields:

[latex]C_2 = \frac{0.850 \;\text{L} \times 5.00 \frac{\text{mol}}{\text{L}}}{one.80 \;\text{L}} = 2.36 \;G[/latex]

This result compares well to our ballpark judge (information technology's a bit less than one-half the stock concentration, five One thousand).

Cheque Your Learning
What is the concentration of the solution that results from diluting 25.0 mL of a 2.04-Thousand solution of CH_iiiOH to 500.0 mL?

Example 7

Book of a Diluted Solution
What book of 0.12 M HBr can be prepared from eleven mL (0.011 50) of 0.45 Yard HBr?

Solution
We are given the book and concentration of a stock solution, 5 _ane and C _i, and the concentration of the resultant diluted solution, C ₂. We need to notice the book of the diluted solution, 5 ₂. We thus rearrange the dilution equation in order to isolate V _ii:

[latex]C_1V_1 = C_2V_2[/latex]
[latex]V_2 = \frac{C_1V_1}{C_2}[/latex]

Since the diluted concentration (0.12 M) is slightly more than i-fourth the original concentration (0.45 M), we would expect the volume of the diluted solution to exist roughly four times the original volume, or around 44 mL. Substituting the given values and solving for the unknown volume yields:

[latex]V_2 = \frac{(0.45\;G)(0.011 \;\text{L})}{0.12 \; M}[/latex]
[latex]V_2 = 0.041 \;\text{L}[/latex]

The book of the 0.12-M solution is 0.041 50 (41 mL). The result is reasonable and compares well with our rough gauge.

Cheque Your Learning
A laboratory experiment calls for 0.125 M HNO₃. What volume of 0.125 Thou HNO₃ can be prepared from 0.250 Fifty of i.88 Chiliad HNO₃?

Example 8

Volume of a Full-bodied Solution Needed for Dilution
What volume of ane.59 M KOH is required to ready v.00 L of 0.100 M KOH?

Solution
We are given the concentration of a stock solution, C _i, and the volume and concentration of the resultant diluted solution, V ₂ and C _two. We demand to find the book of the stock solution, V _i. Nosotros thus rearrange the dilution equation in order to isolate V _i:

[latex]C_1V_1 = C_2V_2[/latex]
[latex]V_2 = \frac{C_2V_2}{C_2}[/latex]

Since the concentration of the diluted solution 0.100 1000 is roughly ane-sixteenth that of the stock solution (1.59 M), we would expect the volume of the stock solution to be about one-sixteenth that of the diluted solution, or around 0.three liters. Substituting the given values and solving for the unknown volume yields:

[latex]V_1 = \frac{(0.100\;Chiliad)(5.00 \;\text{L})}{i.59 \; M}[/latex]
[latex]V_1 = 0.314 \;\text{L}[/latex]

Thus, nosotros would need 0.314 50 of the 1.59-Thou solution to ready the desired solution. This result is consistent with our rough estimate.

Check Your Learning
What book of a 0.575-Grand solution of glucose, C₆H₁₂O_vi, can be prepared from 50.00 mL of a 3.00-G glucose solution?

Key Concepts and Summary

Solutions are homogeneous mixtures. Many solutions comprise ane component, called the solvent, in which other components, chosen solutes, are dissolved. An aqueous solution is 1 for which the solvent is water. The concentration of a solution is a measure of the relative amount of solute in a given corporeality of solution. Concentrations may be measured using diverse units, with 1 very useful unit being molarity, defined as the number of moles of solute per liter of solution. The solute concentration of a solution may be decreased past adding solvent, a procedure referred to as dilution. The dilution equation is a elementary relation between concentrations and volumes of a solution before and later on dilution.

Cardinal Equations

• [latex]M = \frac{\text{mol solute}}{\text{L solution}}[/latex]
• C _i 5 ₁ = C ₂ V _ii

Chemistry Cease of Chapter Exercises

1. Explicate what changes and what stays the same when 1.00 L of a solution of NaCl is diluted to 1.80 L.
2. What information do we need to calculate the molarity of a sulfuric acid solution?
3. What does it hateful when nosotros say that a 200-mL sample and a 400-mL sample of a solution of salt have the same molarity? In what ways are the two samples identical? In what ways are these 2 samples different?
4. Make up one's mind the molarity for each of the following solutions:

   (a) 0.444 mol of CoCl_two in 0.654 L of solution

   (b) 98.0 chiliad of phosphoric acid, H_threePO₄, in i.00 Fifty of solution

   (c) 0.2074 g of calcium hydroxide, Ca(OH)₂, in 40.00 mL of solution

   (d) x.5 kg of Na₂SO_four·10H₂O in 18.threescore 50 of solution

   (eastward) seven.0 × 10⁻³ mol of I₂ in 100.0 mL of solution

   (f) one.eight × 10^iv mg of HCl in 0.075 L of solution

5. Determine the molarity of each of the post-obit solutions:

   (a) i.457 mol KCl in 1.500 L of solution

   (b) 0.515 g of H₂So₄ in i.00 L of solution

   (c) 20.54 g of Al(NO₃)_iii in 1575 mL of solution

   (d) two.76 kg of CuSO_four·5H_iiO in 1.45 L of solution

   (e) 0.005653 mol of Br₂ in 10.00 mL of solution

   (f) 0.000889 g of glycine, C₂H₅NO_two, in 1.05 mL of solution

6. Consider this question: What is the mass of the solute in 0.500 L of 0.thirty M glucose, C_{half dozen}H₁₂O₆, used for intravenous injection?

   (a) Outline the steps necessary to answer the question.

   (b) Answer the question.

7. Consider this question: What is the mass of solute in 200.0 L of a 1.556-One thousand solution of KBr?

   (a) Outline the steps necessary to answer the question.

   (b) Respond the question.

8. Summate the number of moles and the mass of the solute in each of the following solutions:

   (a) 2.00 Fifty of xviii.5 Grand H_iiSO₄, concentrated sulfuric acid

   (b) 100.0 mL of 3.8 × 10⁻⁵ M NaCN, the minimum lethal concentration of sodium cyanide in blood serum

   (c) 5.l L of 13.3 M H₂CO, the formaldehyde used to "set up" tissue samples

   (d) 325 mL of 1.8 × x⁻⁶ M FeSO₄, the minimum concentration of iron sulfate detectable past gustatory modality in drinking h2o

9. Calculate the number of moles and the mass of the solute in each of the post-obit solutions:

   (a) 325 mL of 8.23 × 10⁻⁵ M KI, a source of iodine in the diet

   (b) 75.0 mL of 2.2 × x⁻⁵ M H₂And then₄, a sample of acrid rain

   (c) 0.2500 Fifty of 0.1135 M K₂CrO_four, an analytical reagent used in atomic number 26 assays

   (d) 10.v L of iii.716 M (NH₄)₂SO₄, a liquid fertilizer

10. Consider this question: What is the molarity of KMnO₄ in a solution of 0.0908 g of KMnO₄ in 0.500 L of solution?

    (a) Outline the steps necessary to answer the question.

    (b) Respond the question.

11. Consider this question: What is the molarity of HCl if 35.23 mL of a solution of HCl incorporate 0.3366 chiliad of HCl?

    (a) Outline the steps necessary to answer the question.

    (b) Answer the question.

12. Calculate the molarity of each of the post-obit solutions:

    (a) 0.195 g of cholesterol, C₂₇H₄₆O, in 0.100 L of serum, the boilerplate concentration of cholesterol in human being serum

    (b) four.25 thou of NH_iii in 0.500 L of solution, the concentration of NH₃ in household ammonia

    (c) 1.49 kg of isopropyl alcohol, C₃H₇OH, in 2.50 L of solution, the concentration of isopropyl alcohol in rubbing alcohol

    (d) 0.029 g of I₂ in 0.100 L of solution, the solubility of I₂ in water at 20 °C

13. Calculate the molarity of each of the following solutions:

    (a) 293 g HCl in 666 mL of solution, a concentrated HCl solution

    (b) two.026 g FeCl₃ in 0.1250 L of a solution used as an unknown in general chemistry laboratories

    (c) 0.001 mg Cdⁱⁱ⁺ in 0.100 L, the maximum permissible concentration of cadmium in drinking water

    (d) 0.0079 g C₇H_vSNO₃ in one ounce (29.6 mL), the concentration of saccharin in a nutrition soft drinkable.

14. In that location is nearly 1.0 g of calcium, equally Ca²⁺, in 1.0 L of milk. What is the molarity of Ca²⁺ in milk?
15. What volume of a 1.00-M Atomic number 26(NO_iii)_iii solution can exist diluted to prepare i.00 L of a solution with a concentration of 0.250 M?
16. If 0.1718 L of a 0.3556-G C₃H₇OH solution is diluted to a concentration of 0.1222 M, what is the volume of the resulting solution?
17. If 4.12 L of a 0.850 Thou-H₃PO_iv solution is be diluted to a volume of 10.00 L, what is the concentration of the resulting solution?
18. What volume of a 0.33-M C₁₂H₂₂O₁₁ solution tin be diluted to prepare 25 mL of a solution with a concentration of 0.025 Chiliad?
19. What is the concentration of the NaCl solution that results when 0.150 L of a 0.556-Thousand solution is immune to evaporate until the volume is reduced to 0.105 L?
20. What is the molarity of the diluted solution when each of the following solutions is diluted to the given concluding book?

    (a) ane.00 Fifty of a 0.250-Grand solution of Fe(NO₃)₃ is diluted to a final volume of 2.00 L

    (b) 0.5000 L of a 0.1222-K solution of C_iiiH_viiOH is diluted to a terminal book of one.250 50

    (c) 2.35 L of a 0.350-M solution of H₃PO₄ is diluted to a final volume of 4.00 50

    (d) 22.l mL of a 0.025-M solution of C₁₂H₂₂O₁₁ is diluted to 100.0 mL

21. What is the concluding concentration of the solution produced when 225.5 mL of a 0.09988-M solution of Na₂CO_three is immune to evaporate until the solution volume is reduced to 45.00 mL?
22. A 2.00-Fifty canteen of a solution of concentrated HCl was purchased for the general chemistry laboratory. The solution contained 868.eight thou of HCl. What is the molarity of the solution?
23. An experiment in a general chemical science laboratory calls for a 2.00-Chiliad solution of HCl. How many mL of eleven.9 Chiliad HCl would be required to make 250 mL of 2.00 Thou HCl?
24. What volume of a 0.xx-M K₂SO_iv solution contains 57 k of K_twoSO₄?
25. The US Environmental Protection Agency (EPA) places limits on the quantities of toxic substances that may exist discharged into the sewer organization. Limits have been established for a diverseness of substances, including hexavalent chromium, which is limited to 0.50 mg/L. If an industry is discharging hexavalent chromium as potassium dichromate (1000₂Cr_iiO₇), what is the maximum permissible molarity of that substance?

Glossary

aqueous solution

solution for which water is the solvent

concentrated

qualitative term for a solution containing solute at a relatively high concentration

concentration

quantitative measure of the relative amounts of solute and solvent nowadays in a solution

dilute

qualitative term for a solution containing solute at a relatively low concentration

dilution

process of adding solvent to a solution in order to lower the concentration of solutes

dissolved

describes the procedure by which solute components are dispersed in a solvent

molarity (M)

unit of concentration, defined every bit the number of moles of solute dissolved in one liter of solution

solute

solution component present in a concentration less than that of the solvent

solvent

solution component present in a concentration that is higher relative to other components

Solutions

Answers to Chemistry End of Chapter Exercises

2. We need to know the number of moles of sulfuric acrid dissolved in the solution and the book of the solution.

four. (a) 0.679 Thousand;
(b) 1.00 M;
(c) 0.06998 M;
(d) one.75 M;
(e) 0.070 G;
(f) 6.half dozen M

half-dozen. (a) determine the number of moles of glucose in 0.500 L of solution; determine the molar mass of glucose; determine the mass of glucose from the number of moles and its molar mass; (b) 27 g

8. (a) 37.0 mol H₂Then₄;
3.63 × 10³ m H_twoSO_four;
(b) iii.8 × ten⁻⁶ mol NaCN;
1.9 × 10^−iv 1000 NaCN;
(c) 73.two mol H₂CO;
2.xx kg H₂CO;
(d) v.9 × 10^−vii mol FeSO_four;
eight.9 × 10^−five yard FeSO_four

10. (a) Determine the molar mass of KMnO₄; determine the number of moles of KMnO₄ in the solution; from the number of moles and the volume of solution, decide the molarity; (b) 1.15 × ten⁻³ 1000

12. (a) 5.04 × 10⁻³ M;
(b) 0.499 Chiliad;
(c) ix.92 M;
(d) i.1 × x⁻³ M

14. 0.025 Yard

16. 0.5000 L

18. one.9 mL

20. (a) 0.125 M;
(b) 0.04888 M;
(c) 0.206 Yard;
(e) 0.0056 M

22. 11.ix Thousand

24. ane.6 50

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Source: https://opentextbc.ca/chemistry/chapter/3-3-molarity/

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