what volume of 1.00 m hcl is needed to prepare 300ml of 0.250m hcl

Chapter iii. Limerick of Substances and Solutions

3.3 Molarity

Learning Objectives

By the end of this section, yous will be able to:

• Describe the fundamental backdrop 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 thing that comprise only one type of chemical element or chemical compound. Even so, mixtures—samples of matter containing 2 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 backdrop. The relative amount of oxygen in a planet'southward 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 forcefulness and resistance to corrosion. The relative corporeality of the active ingredient in a medicine determines its effectiveness in achieving the desired pharmacological result. The relative amount of sugar in a beverage determines its sweetness (encounter Effigy 1). In this section, we will describe i of the most mutual ways in which the relative compositions of mixtures may be quantified.

Effigy 1. Sugar is i of many components in the complex mixture known as java. The corporeality of saccharide in a given corporeality of java is an important determinant of the beverage'due south sugariness. (credit: Jane Whitney)

Solutions

We have previously defined solutions equally homogeneous mixtures, pregnant that the composition of the mixture (and therefore its properties) is uniform throughout its unabridged volume. Solutions occur oftentimes in nature and take also been implemented in many forms of manmade technology. We will explore a more thorough handling of solution properties in the affiliate on solutions and colloids, only here we will innovate some of the basic properties of solutions.

The relative amount of a given solution component is known as its concentration. Ofttimes, though not always, a solution contains one component with a concentration that is significantly greater than that of all other components. This component is called the solvent and may be viewed every bit the medium in which the other components are dispersed, or dissolved. Solutions in which water is the solvent are, of class, very common on our planet. A solution in which water 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 every bit dilute (of relatively low concentration) and concentrated (of relatively high concentration).

Concentrations may be quantitatively assessed using a broad diversity of measurement units, each convenient for particular applications. Molarity (M) is a useful concentration unit for many applications in chemistry. Molarity is defined as the number of moles of solute in exactly 1 liter (1 L) of the solution:

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

Example 1

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

Solution
Since the molar 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]K = \frac{\text{mol solute}}{\text{L solution}} = \frac{0.133 \;\text{mol}}{355 \;\text{mL} \times \frac{ane \;\text{L}}{1000 \;\text{mL}}} = 0.375 \; M[/latex]

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

Instance 2

Deriving Moles and Volumes from Molar Concentrations
How much saccharide (mol) is contained in a small-scale sip (~x mL) of the soft beverage from Example one?

Solution
In this instance, we tin can rearrange the definition of molarity to isolate the quantity sought, moles of sugar. We so substitute the value for molarity that we derived in Example i, 0.375 M:

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

[latex]\text{mol solute} = 0.375 \;\frac{\text{mol saccharide}}{\text{L}} \times (10 \;\text{mL} \times \frac{ane \text{50}}{grand \;\text{mL}}) = 0.004 \;\text{mol sugar}[/latex]

Cheque Your Learning
What volume (mL) of the sweetened tea described in Case one contains the aforementioned amount of sugar (mol) as 10 mL of the soft drink in this example?

Example iii

Calculating Molar Concentrations from the Mass of Solute
Distilled white vinegar (Figure two) is a solution of acerb acid, CH₃CO₂H, in water. A 0.500-L vinegar solution contains 25.ii g of acetic acid. What is the concentration of the acerb acid solution in units of molarity?

Figure 2. Distilled white vinegar is a solution of acetic acrid in h2o.

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

[latex]Thousand = \frac{\text{mol solute}}{\text{50 solution}} = \frac{25.2 \;\text{g CH}_3\text{CO}_2\text{H} \times \frac{1 \;\text{mol CH}_2\text{CO}_2\text{H}}{threescore.052 \;\text{g CH}_2\text{CO}_2\text{H}}}{0.500 \;\text{L solution}} = 0.839 \;K[/latex]

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

Check Your Learning
Summate the molarity of half-dozen.52 g of CoCl₂ (128.9 one thousand/mol) dissolved in an aqueous solution with a total volume of 75.0 mL.

Example 4

Determining the Mass of Solute in a Given Book of Solution
How many grams of NaCl are contained in 0.250 Fifty of a 5.30-M solution?

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

[latex]M = \;\frac{\text{mol solute}}{\text{50 solution}}[/latex]
[latex]\text{mol solute} = M \times \text{L solution}[/latex]
[latex]\text{mol solute} = 5.thirty \;\frac{\text{mol NaCl}}{\text{Fifty}} \times 0.250 \;\text{L} = 1.325 \;\text{mol NaCl}[/latex]

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

[latex]one.325 \;\text{mol NaCl} \times \frac{58.44 \;\text{g NaCl}}{\text{mol NaCl}} = 77.iv \;\text{chiliad NaCl}[/latex]

Cheque Your Learning
How many grams of CaCl₂ (110.98 g/mol) are contained in 250.0 mL of a 0.200-M solution of calcium chloride?

When performing calculations stepwise, equally in Example iv, information technology is important to refrain from rounding whatever intermediate calculation results, which tin can lead to rounding errors in the final result. In Example 4, the molar amount of NaCl computed in the first footstep, 1.325 mol, would be properly rounded to i.32 mol if it were to be reported; however, although the last digit (5) is non pregnant, information technology must be retained every bit a baby-sit digit in the intermediate calculation. If we had not retained this guard digit, the last calculation for the mass of NaCl would have been 77.one chiliad, a difference of 0.3 g.

In add-on to retaining a guard digit for intermediate calculations, we tin can likewise avert rounding errors by performing computations in a unmarried stride (come across Case v). This eliminates intermediate steps and then that only the final issue is rounded.

Example five

Determining the Volume of Solution Containing a Given Mass of Solute
In Case three, we found the typical concentration of vinegar to be 0.839 M. What book of vinegar contains 75.six g of acetic acid?

Solution
Commencement, use the tooth mass to calculate moles of acetic acid from the given mass:

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

Then, use the molarity of the solution to summate the volume of solution containing this molar corporeality of solute:

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

Combining these two steps into one yields:

[latex]\text{k solute} \times \frac{\text{mol solute}}{\text{1000 solute}} \times \frac{\text{L solution}}{\text{mol solute}} = \text{L solution}[/latex][latex]75.6 \;\text{1000 CH}_3\text{CO}_2\text{H} (\frac{\text{mol CH}_3\text{CO}_2\text{H}}{60.05 \;\text{chiliad}}) (\frac{\text{L solution}}{0.839 \;\text{mol CH}_3\text{CO}_2\text{H}}) = 1.l \;\text{L solution}[/latex]

Check Your Learning
What volume of a 1.50-M KBr solution contains 66.0 chiliad KBr?

Dilution of Solutions

Dilution is the procedure whereby the concentration of a solution is lessened past the improver of solvent. For instance, 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).

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

Dilution is likewise a common ways of preparing solutions of a desired concentration. By adding solvent to a measured portion of a more concentrated stock solution, we can achieve a particular concentration. For example, commercial pesticides are typically sold as solutions in which the active ingredients are far more than concentrated than is appropriate for their awarding. Earlier they tin be used on crops, the pesticides must be diluted. This is as well a very common exercise for the preparation of a number of common laboratory reagents (Figure 4).

Figure 4. A solution of KMnO₄ is prepared by mixing water with iv.74 thou of KMnO_four in a flask. (credit: modification of work past Marking Ott)

A unproblematic mathematical relationship can be used to relate the volumes and concentrations of a solution earlier and later the dilution process. Co-ordinate to the definition of molarity, the molar amount of solute in a solution is equal to the product of the solution's molarity and its volume in liters:

[latex]due north = ML[/latex]

Expressions like these may be written for a solution before and after it is diluted:

[latex]n_1 = M_1L_1[/latex]

[latex]n_2 = M_2L_2[/latex]

where the subscripts "ane" and "ii" refer to the solution before and after the dilution, respectively. Since the dilution process does non alter the amount of solute in the solution, n ₁ = n _ii. Thus, these two equations may be fix equal to one some other:

[latex]M_1L_1 = M_2L_2[/latex]

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

[latex]C_1V_1 = C_2V_2[/latex]

where C and Five are concentration and book, respectively.

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

Instance half-dozen

Determining the Concentration of a Diluted Solution
If 0.850 L of a five.00-Chiliad solution of copper nitrate, Cu(NO_three)₂, is diluted to a volume of 1.fourscore Fifty by the improver of water, what is the molarity of the diluted solution?

Solution
We are given the volume and concentration of a stock solution, V _one and C ₁, and the volume of the resultant diluted solution, V ₂. We need to find the concentration of the diluted solution, C ₂. Nosotros thus rearrange the dilution equation in social club 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 beingness diluted past more than two-fold (book is increased from 0.85 L to i.80 L), we would expect the diluted solution'south concentration to be less than half 5 M. We will compare this ballpark estimate to the calculated result to check for whatsoever gross errors in computation (for instance, such as 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{50}}}{i.fourscore \;\text{L}} = 2.36 \;1000[/latex]

This consequence compares well to our ballpark estimate (information technology'southward a bit less than ane-half the stock concentration, five G).

Check Your Learning
What is the concentration of the solution that results from diluting 25.0 mL of a ii.04-M solution of CH₃OH to 500.0 mL?

Case vii

Volume of a Diluted Solution
What volume of 0.12 M HBr can exist prepared from 11 mL (0.011 L) of 0.45 Thou HBr?

Solution
We are given the volume and concentration of a stock solution, V _one and C ₁, and the concentration of the resultant diluted solution, C ₂. We need to find the book of the diluted solution, V ₂. We thus rearrange the dilution equation in order to isolate V _two:

[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 1-fourth the original concentration (0.45 M), we would expect the volume of the diluted solution to be roughly four times the original volume, or around 44 mL. Substituting the given values and solving for the unknown book yields:

[latex]V_2 = \frac{(0.45\;One thousand)(0.011 \;\text{L})}{0.12 \; Yard}[/latex]
[latex]V_2 = 0.041 \;\text{L}[/latex]

The book of the 0.12-Yard solution is 0.041 L (41 mL). The result is reasonable and compares well with our crude estimate.

Check Your Learning
A laboratory experiment calls for 0.125 Thousand HNO₃. What book of 0.125 M HNO₃ tin can be prepared from 0.250 L of one.88 Yard HNO_three?

Instance 8

Volume of a Concentrated Solution Needed for Dilution
What volume of 1.59 M KOH is required to prepare 5.00 L of 0.100 K KOH?

Solution
We are given the concentration of a stock solution, C ₁, and the volume and concentration of the resultant diluted solution, 5 ₂ and C ₂. We demand to find the volume of the stock solution, V _one. We thus rearrange the dilution equation in order to isolate 5 _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 M is roughly one-sixteenth that of the stock solution (1.59 1000), we would expect the volume of the stock solution to be about 1-sixteenth that of the diluted solution, or effectually 0.3 liters. Substituting the given values and solving for the unknown volume yields:

[latex]V_1 = \frac{(0.100\;K)(5.00 \;\text{L})}{1.59 \; M}[/latex]
[latex]V_1 = 0.314 \;\text{Fifty}[/latex]

Thus, nosotros would need 0.314 L of the 1.59-M solution to prepare the desired solution. This result is consistent with our crude estimate.

Cheque Your Learning
What volume of a 0.575-M solution of glucose, C₆H₁₂O₆, tin be prepared from 50.00 mL of a iii.00-M glucose solution?

Cardinal Concepts and Summary

Solutions are homogeneous mixtures. Many solutions incorporate one component, chosen the solvent, in which other components, called solutes, are dissolved. An aqueous solution is one 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 various units, with i 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 by calculation solvent, a process referred to as dilution. The dilution equation is a uncomplicated relation between concentrations and volumes of a solution before and afterwards dilution.

Cardinal Equations

• [latex]Chiliad = \frac{\text{mol solute}}{\text{L solution}}[/latex]
• C ₁ V _ane = C ₂ V ₂

Chemistry Finish of Chapter Exercises

1. Explain what changes and what stays the aforementioned when 1.00 L of a solution of NaCl is diluted to 1.80 L.
2. What information do we need to summate the molarity of a sulfuric acid solution?
3. What does it mean when we say that a 200-mL sample and a 400-mL sample of a solution of common salt have the same molarity? In what ways are the two samples identical? In what means are these ii samples different?
4. Decide the molarity for each of the following solutions:

   (a) 0.444 mol of CoCl₂ in 0.654 L of solution

   (b) 98.0 grand of phosphoric acid, H_threePO₄, in one.00 L of solution

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

   (d) 10.five kg of Na₂And so_iv·10H_iiO in 18.60 50 of solution

   (east) 7.0 × ten⁻³ mol of I₂ in 100.0 mL of solution

   (f) 1.8 × 10⁴ mg of HCl in 0.075 L of solution

5. Make up one's mind the molarity of each of the post-obit solutions:

   (a) 1.457 mol KCl in ane.500 L of solution

   (b) 0.515 g of H₂And then₄ in 1.00 Fifty of solution

   (c) 20.54 g of Al(NO₃)₃ in 1575 mL of solution

   (d) 2.76 kg of CuSO₄·5H₂O in ane.45 L of solution

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

   (f) 0.000889 one thousand of glycine, C₂H₅NO₂, in ane.05 mL of solution

6. Consider this question: What is the mass of the solute in 0.500 L of 0.30 M glucose, C_sixH₁₂O_vi, used for intravenous injection?

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

   (b) Answer the question.

7. Consider this question: What is the mass of solute in 200.0 L of a ane.556-Grand solution of KBr?

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

   (b) Reply the question.

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

   (a) 2.00 50 of 18.5 M H_twoSO_iv, full-bodied sulfuric acid

   (b) 100.0 mL of 3.eight × ten⁻⁵ One thousand NaCN, the minimum lethal concentration of sodium cyanide in blood serum

   (c) 5.50 50 of 13.3 K H₂CO, the formaldehyde used to "fix" tissue samples

   (d) 325 mL of one.8 × 10⁻⁶ M FeSO₄, the minimum concentration of atomic number 26 sulfate detectable by taste in drinking water

9. Calculate the number of moles and the mass of the solute in each of the following solutions:

   (a) 325 mL of eight.23 × 10⁻⁵ One thousand KI, a source of iodine in the diet

   (b) 75.0 mL of 2.2 × 10⁻⁵ M H_twoSo₄, a sample of acid rain

   (c) 0.2500 L of 0.1135 M K_twoCrO₄, an belittling reagent used in iron assays

   (d) 10.v Fifty of 3.716 M (NH₄)_iiSO₄, a liquid fertilizer

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

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

    (b) Reply the question.

11. Consider this question: What is the molarity of HCl if 35.23 mL of a solution of HCl contain 0.3366 g 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 1000 of cholesterol, C₂₇H₄₆O, in 0.100 Fifty of serum, the average concentration of cholesterol in human serum

    (b) four.25 g of NH₃ in 0.500 50 of solution, the concentration of NH_iii in household ammonia

    (c) one.49 kg of isopropyl booze, C_threeH₇OH, in two.50 L of solution, the concentration of isopropyl alcohol in rubbing alcohol

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

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

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

    (b) two.026 grand FeCl₃ in 0.1250 Fifty of a solution used every bit an unknown in general chemistry laboratories

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

    (d) 0.0079 g C_viiH_vSNO₃ in ane ounce (29.half-dozen mL), the concentration of saccharin in a diet soft potable.

14. There is virtually 1.0 g of calcium, as Ca²⁺, in 1.0 L of milk. What is the molarity of Ca²⁺ in milk?
15. What volume of a 1.00-M Fe(NO_iii)₃ solution can be diluted to prepare 1.00 50 of a solution with a concentration of 0.250 1000?
16. If 0.1718 L of a 0.3556-Thou C₃H_viiOH solution is diluted to a concentration of 0.1222 M, what is the book of the resulting solution?
17. If 4.12 L of a 0.850 Thousand-H_threePO₄ solution is be diluted to a book of 10.00 L, what is the concentration of the resulting solution?
18. What book of a 0.33-M C₁₂H₂₂O_xi solution tin can exist diluted to set 25 mL of a solution with a concentration of 0.025 M?
19. What is the concentration of the NaCl solution that results when 0.150 L of a 0.556-M solution is allowed 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 final volume?

    (a) 1.00 L of a 0.250-M solution of Fe(NO_three)₃ is diluted to a concluding book of ii.00 L

    (b) 0.5000 Fifty of a 0.1222-M solution of C₃H_viiOH is diluted to a final volume of 1.250 L

    (c) 2.35 L of a 0.350-Thou solution of H_threePO_iv is diluted to a final volume of iv.00 L

    (d) 22.50 mL of a 0.025-M solution of C₁₂H₂₂O₁₁ is diluted to 100.0 mL

21. What is the final concentration of the solution produced when 225.5 mL of a 0.09988-M solution of Na₂CO₃ is immune to evaporate until the solution volume is reduced to 45.00 mL?
22. A ii.00-Fifty bottle of a solution of concentrated HCl was purchased for the general chemistry laboratory. The solution contained 868.viii g of HCl. What is the molarity of the solution?
23. An experiment in a general chemistry laboratory calls for a 2.00-M solution of HCl. How many mL of xi.ix M HCl would be required to make 250 mL of 2.00 Thou HCl?
24. What book of a 0.20-M K₂SO₄ solution contains 57 g of K₂SO_four?
25. The US Environmental Protection Agency (EPA) places limits on the quantities of toxic substances that may exist discharged into the sewer system. Limits have been established for a diverseness of substances, including hexavalent chromium, which is limited to 0.l mg/L. If an manufacture is discharging hexavalent chromium as potassium dichromate (K_iiCr₂O₇), what is the maximum permissible molarity of that substance?

Glossary

aqueous solution

solution for which h2o is the solvent

full-bodied

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

concentration

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

dilute

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

dilution

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

dissolved

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

molarity (Chiliad)

unit of concentration, defined equally the number of moles of solute dissolved in i liter of solution

solute

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

solvent

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

Solutions

Answers to Chemistry Stop of Affiliate Exercises

2. We need to know the number of moles of sulfuric acid dissolved in the solution and the volume of the solution.

4. (a) 0.679 M;
(b) 1.00 M;
(c) 0.06998 M;
(d) 1.75 M;
(due east) 0.070 Thou;
(f) half dozen.6 M

6. (a) make up one's mind 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 thou

8. (a) 37.0 mol H₂So₄;
3.63 × 10ⁱⁱⁱ g H_iiAnd so_four;
(b) iii.eight × 10⁻⁶ mol NaCN;
1.9 × 10⁻⁴ grand NaCN;
(c) 73.2 mol H₂CO;
2.20 kg H₂CO;
(d) five.9 × 10^−seven mol FeSO₄;
eight.nine × 10⁻⁵ g FeSO₄

10. (a) Decide the molar mass of KMnO₄; decide the number of moles of KMnO₄ in the solution; from the number of moles and the volume of solution, determine the molarity; (b) ane.15 × 10⁻³ Yard

12. (a) 5.04 × ten⁻³ M;
(b) 0.499 M;
(c) 9.92 M;
(d) 1.1 × ten⁻³ M

14. 0.025 One thousand

16. 0.5000 L

eighteen. 1.nine mL

20. (a) 0.125 1000;
(b) 0.04888 M;
(c) 0.206 M;
(eastward) 0.0056 Thou

22. xi.9 One thousand

24. 1.6 L

silvercoveregs.blogspot.com

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

Share this post