How Do You Know if Hcl(Aq) Is Present in a Solution

ix.two Concentration

Learning Objectives

1. Express the amount of solute in a solution in diverse concentration units.
2. Utilise molarity to determine quantities in chemical reactions.
3. Determine the resulting concentration of a diluted solution.

To define a solution precisely, nosotros demand to country its concentrationHow much solute is dissolved in a certain amount of solvent. : how much solute is dissolved in a certain amount of solvent. Words such equally dilute or concentrated are used to describe solutions that take a lilliputian or a lot of dissolved solute, respectively, only these are relative terms whose meanings depend on various factors.

Solubility

In that location is usually a limit to how much solute will dissolve in a given amount of solvent. This limit is called the solubilityThe limit of how much solute tin exist dissolved in a given amount of solvent. of the solute. Some solutes accept a very small solubility, while other solutes are soluble in all proportions. Table 9.ii "Solubilities of Diverse Solutes in Water at 25°C (Except as Noted)" lists the solubilities of various solutes in water. Solubilities vary with temperature, and then Table 9.two "Solubilities of Various Solutes in Water at 25°C (Except equally Noted)" includes the temperature at which the solubility was adamant.

Table ix.two Solubilities of Various Solutes in Water at 25°C (Except as Noted)

Substance	Solubility (m in 100 mL of H2O)
AgCl(s)	0.019
CviH6(ℓ) (benzene)	0.178
CH4(one thousand)	0.0023
CO2(chiliad)	0.150
CaCOiii(s)	0.058
CaF2(s)	0.0016
Ca(NOiii)two(south)	143.9
C6H12Ovi (glucose)	120.3 (at thirty°C)
KBr(s)	67.8
MgCO3(s)	ii.20
NaCl(due south)	36.0
NaHCO3(s)	viii.41
C12H22O11 (sucrose)	204.0 (at 20°C)

If a solution contains so much solute that its solubility limit is reached, the solution is said to be saturatedA solution whose solute is at its solubility limit. , and its concentration is known from data contained in Table 9.2 "Solubilities of Diverse Solutes in H2o at 25°C (Except as Noted)". If a solution contains less solute than the solubility limit, it is unsaturatedA solution whose solute is less than its solubility limit. . Nether special circumstances, more than solute can exist dissolved even after the normal solubility limit is reached; such solutions are called supersaturated and are not stable. If the solute is solid, excess solute tin hands recrystallize. If the solute is a gas, it tin can chimera out of solution uncontrollably, like what happens when y'all milk shake a soda can and so immediately open up it.

Notation

Recrystallization of excess solute from a supersaturated solution usually gives off energy as oestrus. Commercial heat packs containing supersaturated sodium acetate (NaC₂H₃O_two) take advantage of this phenomenon. You tin can probably find them at your local drugstore.

Almost solutions we encounter are unsaturated, so knowing the solubility of the solute does non accurately limited the amount of solute in these solutions. There are several common ways of specifying the concentration of a solution.

Percent Composition

There are several ways of expressing the concentration of a solution by using a percentage. The mass/mass percentA concentration unit of measurement that relates the mass of the solute to the mass of the solution. (% chiliad/m) is divers as the mass of a solute divided by the mass of a solution times 100:

$$\%\text{m/yard =}\frac{\text{mass of solute}}{\text{mass of solution}}\times 100\%$$

If you tin measure the masses of the solute and the solution, determining the mass/mass percent is easy. Each mass must exist expressed in the same units to determine the proper concentration.

Example 2

A saline solution with a mass of 355 k has 36.5 g of NaCl dissolved in information technology. What is the mass/mass percent concentration of the solution?

Solution

We can substitute the quantities given in the equation for mass/mass percent:

$$\%\text{ m/g}=\frac{\text{36}\text{.v thou}}{\text{355 g}}\times \mathrm{100\%}=\text{ten}\text{.3\%}$$

Skill-Building Exercise

1. A dextrose (as well chosen D-glucose, C_viH₁₂O₆) solution with a mass of 2.00 × 10² thousand has 15.8 g of dextrose dissolved in information technology. What is the mass/mass percent concentration of the solution?

For gases and liquids, volumes are relatively easy to measure, so the concentration of a liquid or a gas solution can be expressed as a volume/volume percentA concentration unit of measurement that relates the volume of the solute to the volume of the solution. (% five/v): the volume of a solute divided by the book of a solution times 100:

$$\text{\% v/v =}\frac{\text{book of solute}}{\text{volume of solution}}\times 100\%$$

Again, the units of the solute and the solution must be the same. A hybrid concentration unit, mass/volume per centumA concentration unit that relates the mass of the solute to the volume of the solution. (% m/five), is unremarkably used for intravenous (Four) fluids (Figure 9.2 "Mass/Volume Percent"). It is defined as the mass in grams of a solute, divided by volume in milliliters of solution times 100:

$$\text{\% one thousand/v =}\frac{\text{mass of solute (thousand)}}{\text{volume of solution (mL)}}\times 100\%$$

Each percent concentration can be used to produce a conversion cistron between the corporeality of solute, the amount of solution, and the percent. Furthermore, given any two quantities in any percent composition, the 3rd quantity tin be calculated, as the following example illustrates.

Instance 3

A sample of 45.0% v/v solution of ethanol (C₂H₅OH) in water has a volume of 115 mL. What volume of ethanol solute does the sample contain?

Solution

A pct concentration is simply the number of parts of solute per 100 parts of solution. Thus, the percent concentration of 45.0% 5/v implies the following:

$$\text{45}.0\%\text{ v/v}\to \frac{45{\text{ mL C}}_{\text{2}}{\text{H}}_5\text{OH}}{\text{100 mL solution}}$$

That is, there are 45 mL of C₂H₅OH for every 100 mL of solution. We tin can apply this fraction as a conversion cistron to decide the corporeality of C₂H₅OH in 115 mL of solution:

$$115\cancel{\text{ mL solution}}\times \frac{{\text{45 mL C}}_{\text{2}}{\text{H}}_5\text{OH}}{\text{100}\cancel{\text{ mL solution}}}=51.8{\text{ mL C}}_{\text{ii}}{\text{H}}_5\text{OH}$$

Note

The highest concentration of ethanol that tin be obtained normally is 95% ethanol, which is really 95% v/v.

Skill-Edifice Do

1. What volume of a 12.75% yard/v solution of glucose (C₆H₁₂O₆) in h2o is needed to obtain 50.0 g of C₆H₁₂O_six?

Case 4

A normal saline Iv solution contains 9.0 g of NaCl in every liter of solution. What is the mass/book percent of normal saline?

Solution

We tin use the definition of mass/volume percent, merely first nosotros have to express the volume in milliliter units:

ane Fifty = ane,000 mL

Considering this is an exact relationship, it does not affect the meaning figures of our upshot.

$$\%\text{ m/v =}\frac{\text{9}\text{.0 g NaCl}}{\text{one,000 mL solution}}\times 100\%=0.90\%\text{ thousand/v}$$

Skill-Building Practise

1. The chlorine bleach that y'all might find in your laundry room is typically equanimous of 27.0 g of sodium hypochlorite (NaOCl), dissolved to make 500.0 mL of solution. What is the mass/book percentage of the bleach?

In improver to percentage units, the units for expressing the concentration of extremely dilute solutions are parts per million (ppm)The mass of a solute compared to the mass of a solution times 1,000,000. and parts per billion (ppb)The mass of a solute compared to the mass of a solution times ane,000,000,000. . Both of these units are mass based and are defined as follows:

$$\text{ppm}=\frac{\text{mass of solute}}{\text{mass of solution}}\times \mathrm{1,000,000}$$ $$\text{ppb}=\frac{\text{mass of solute}}{\text{mass of solution}}\times \mathrm{1,000,000,000}$$

Note

Similar to parts per one thousand thousand and parts per billion, related units include parts per thousand (ppth) and parts per trillion (ppt).

Concentrations of trace elements in the trunk—elements that are present in extremely low concentrations only are all the same necessary for life—are ordinarily expressed in parts per million or parts per billion. Concentrations of poisons and pollutants are also described in these units. For example, cobalt is present in the body at a concentration of 21 ppb, while the State of Oregon'due south Department of Agriculture limits the concentration of arsenic in fertilizers to 9 ppm.

Note

In aqueous solutions, 1 ppm is essentially equal to 1 mg/L, and 1 ppb is equivalent to 1 µg/Fifty.

Example 5

If the concentration of cobalt in a human trunk is 21 ppb, what mass in grams of Co is present in a trunk having a mass of seventy.0 kg?

Solution

A concentration of 21 ppb ways "21 g of solute per 1,000,000,000 g of solution." Written as a conversion factor, this concentration of Co is as follows:

$$\text{21 ppb Co}\to \frac{21\text{ g Co}}{\text{1,000,000,000 g solution}}$$

We can use this as a conversion cistron, simply first we must catechumen 70.0 kg to gram units:

$$\mathrm{seventy.0}\text{ kg}\times \frac{\text{1,000 g}}{\text{one kg}}=7.00\times {10}^4\text{ g}$$

Now we decide the corporeality of Co:

$$\mathrm{seven.00}\times {\mathrm{ten}}^4\cancel{\text{ g solution}}\times \frac{\text{21 one thousand Co}}{\text{1,000,000,000}\cancel{\text{ m solution}}}=0.0015\text{ g Co}$$

This is only i.v mg.

Skill-Building Exercise

1. An 85 kg body contains 0.012 g of Ni. What is the concentration of Ni in parts per million?

Molarity

Another way of expressing concentration is to requite the number of moles of solute per unit volume of solution. Such concentration units are useful for discussing chemical reactions in which a solute is a product or a reactant. Molar mass tin can so be used equally a conversion factor to convert amounts in moles to amounts in grams.

MolarityNumber of moles of solute per liter of solution. is defined as the number of moles of a solute dissolved per liter of solution:

$$\text{molarity}=\frac{\text{number of moles of solute}}{\text{number of liters of solution}}$$

Molarity is abbreviated Grand (often referred to as "molar"), and the units are often abbreviated as mol/L. It is important to remember that "mol" in this expression refers to moles of solute and that "L" refers to liters of solution. For case, if you have 1.5 mol of NaCl dissolved in 0.500 50 of solution, its molarity is therefore

$$\frac{\mathrm{ane.5}\text{ mol NaCl}}{\text{0}\text{.500 L solution}}=3.0\text{ M NaCl}$$

which is read as "three bespeak oh molar sodium chloride." Sometimes (aq) is added when the solvent is water, as in "3.0 M NaCl(aq)."

Before a molarity concentration can be calculated, the amount of the solute must be expressed in moles, and the book of the solution must exist expressed in liters, as demonstrated in the post-obit example.

Example 6

What is the molarity of an aqueous solution of 25.0 g of NaOH in 750 mL?

Solution

Earlier we substitute these quantities into the definition of molarity, we must convert them to the proper units. The mass of NaOH must be converted to moles of NaOH. The molar mass of NaOH is 40.00 g/mol:

$$25.0\cancel{\text{ g NaOH}}\times \frac{\text{i mol NaOH}}{\text{xl}\text{.00}\cancel{\text{ g NaOH}}}=0.625\text{ mol NaOH}$$

Next, we convert the volume units from milliliters to liters:

$$750\cancel{\text{ mL}}\times \frac{\text{1 L}}{\text{1,000}\cancel{\text{ mL}}}=0.750\text{ Fifty}$$

Now that the quantities are expressed in the proper units, nosotros can substitute them into the definition of molarity:

$$\text{M}=\frac{\text{0}\text{.625 mol NaOH}}{\text{0}\text{.750 L}}=0.833\text{ M NaOH}$$

Skill-Edifice Do

1. If a 350 mL loving cup of coffee contains 0.150 thousand of caffeine (C₈H₁₀N_fourO₂), what is the molarity of this caffeine solution?

The definition of molarity can also be used to summate a needed book of solution, given its concentration and the number of moles desired, or the number of moles of solute (and subsequently, the mass of the solute), given its concentration and book. The following example illustrates this.

Example 7

1. What volume of a 0.0753 One thousand solution of dimethylamine [(CH₃)_iiNH] is needed to obtain 0.450 mol of the chemical compound?
2. Ethylene glycol (C₂H_viO₂) is mixed with water to make machine engine coolants. How many grams of C₂H_sixO₂ are in five.00 L of a 6.00 Thou aqueous solution?

Solution

In both parts, nosotros volition use the definition of molarity to solve for the desired quantity.

1. $$0.0753\text{ M}=\frac{\text{0}{\text{.450 mol (CH}}_{\text{three}}{)}_{\mathrm{ii}}\text{NH}}{\text{book of solution}}$$

   To solve for the book of solution, we multiply both sides by volume of solution and carve up both sides past the molarity value to isolate the book of solution on one side of the equation:

   $$\text{book of solution}=\frac{\text{0}{\text{.450 mol (CH}}_{\text{three}}{)}_2\text{NH}}{\text{0}\text{.0753 G}}=\mathrm{v.98}\text{ 50}$$

   Annotation that because the definition of molarity is mol/L, the segmentation of mol by M yields L, a unit of book.

2. The tooth mass of C_iiH_sixO₂ is 62.08 g/mol., and so

   $$6.00\text{ M}=\frac{\text{moles of solute}}{\text{5}\text{.00 L}}$$

   To solve for the number of moles of solute, we multiply both sides past the volume:

   moles of solute = (6.00 G)(v.00 50) = xxx.0 mol

   Note that considering the definition of molarity is mol/Fifty, the product One thousand × L gives mol, a unit of amount. Now, using the tooth mass of C₃H₈O_three, we convert mol to thou:

   $$30.0\cancel{\text{ mol}}\times \frac{\text{62}\text{.08 thou}}{\cancel{\text{ mol}}}=\mathrm{ane,860}\text{ g}$$

   Thus, there are 1,860 m of C_twoH_half-dozenO₂ in the specified amount of engine coolant.

Skill-Building Exercise

1. What book of a 0.0902 M solution of formic acid (HCOOH) is needed to obtain 0.888 mol of HCOOH?

2. Acetic acrid (HC₂H_threeO_two) is the acid in vinegar. How many grams of HC₂H_threeO₂ are in 0.565 L of a 0.955 K solution?

Using Molarity in Stoichiometry Problems

Of all the ways of expressing concentration, molarity is the i most normally used in stoichiometry problems because information technology is directly related to the mole unit. Consider the following chemical equation:

HCl(aq) + NaOH(s) → H₂O(ℓ) + NaCl(aq)

Suppose we want to know how many liters of aqueous HCl solution volition react with a given mass of NaOH. A typical arroyo to answering this question is as follows:

In itself, each step is a straightforward conversion. Information technology is the combination of the steps that is a powerful quantitative tool for problem solving.

Instance 8

How many milliliters of a 2.75 M HCl solution are needed to react with 185 g of NaOH? The balanced chemical equation for this reaction is as follows:

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

Solution

We will follow the flowchart to reply this question. First, nosotros convert the mass of NaOH to moles of NaOH using its molar mass, 40.00 one thousand/mol:

$$185\cancel{\text{ m NaOH}}\times \frac{\text{ane mol NaOH}}{\text{twoscore}\text{.00}\cancel{\text{ g NaOH}}}=4.63\text{ mol NaOH}$$

Using the balanced chemical equation, we see that in that location is a one-to-ane ratio of moles of HCl to moles of NaOH. We use this to decide the number of moles of HCl needed to react with the given corporeality of NaOH:

$$4.63\cancel{\text{ mol NaOH}}\times \frac{\text{1 mol HCl}}{\text{1}\cancel{\text{ mol NaOH}}}=\mathrm{four.63}\text{ mol HCl}$$

Finally, we use the definition of molarity to decide the book of two.75 M HCl needed:

$$2.75\text{ M HCl}=\frac{\text{4}\text{.63 mol HCl}}{\text{book of HCl solution}}$$ $$\text{volume of HCl}=\frac{\text{4}\text{.63 mol HCl}}{\text{2}\text{.75 M HCl}}=\mathrm{i.68}\cancel{\text{ L}}\times \frac{\text{1,000 mL}}{\text{1}\cancel{\text{ 50}}}=\mathrm{one,680}\text{ mL}$$

We need i,680 mL of 2.75 M HCl to react with the NaOH.

Skill-Edifice Practise

1. How many milliliters of a i.04 K H₂So₄ solution are needed to react with 98.five thou of Ca(OH)₂? The balanced chemical equation for the reaction is as follows:

   H_twoAnd then_four(aq) + Ca(OH)₂(s) → 2H_iiO(ℓ) + CaSO₄(aq)

The full general steps for performing stoichiometry issues such equally this are shown in Figure 9.3 "Diagram of Steps for Using Molarity in Stoichiometry Calculations". You may want to consult this figure when working with solutions in chemical reactions. The double arrows in Effigy nine.iii "Diagram of Steps for Using Molarity in Stoichiometry Calculations" betoken that you can start at either end of the nautical chart and, after a series of simple conversions, determine the quantity at the other end.

Effigy 9.3 Diagram of Steps for Using Molarity in Stoichiometry Calculations

When using molarity in stoichiometry calculations, a specific sequence of steps normally leads you to the right answer.

Many of the fluids found in our bodies are solutions. The solutes range from unproblematic ionic compounds to complex proteins. Table nine.3 "Approximate Concentrations of Various Solutes in Some Solutions in the Body*" lists the typical concentrations of some of these solutes.

Table nine.3 Approximate Concentrations of Various Solutes in Some Solutions in the Torso*

Solution	Solute	Concentration (M)
blood plasma	Na+	0.138
	K+	0.005
	Caii+	0.004
	Mg2+	0.003
	Cl−	0.110
	HCO3 −	0.030
breadbasket acid	HCl	0.10
urine	NaCl	0.xv
	PO4 iii−	0.05
	NH2CONHtwo (urea)	0.30
*Annotation: Concentrations are approximate and can vary widely.

Looking Closer: The Dose Makes the Poison

Why is it that nosotros can drink 1 qt of water when nosotros are thirsty and not be harmed, but if nosotros ingest 0.5 g of arsenic, we might die? In that location is an old saying: the dose makes the poison. This ways that what may be unsafe in some amounts may not be dangerous in other amounts.

Take arsenic, for example. Some studies show that arsenic impecuniousness limits the growth of animals such every bit chickens, goats, and pigs, suggesting that arsenic is actually an essential trace chemical element in the nutrition. Humans are constantly exposed to tiny amounts of arsenic from the environment, then studies of completely arsenic-free humans are not bachelor; if arsenic is an essential trace mineral in homo diets, it is probably required on the order of 50 ppb or less. A toxic dose of arsenic corresponds to about 7,000 ppb and college, which is over 140 times the trace amount that may be required by the body. Thus, arsenic is non poisonous in and of itself. Rather, it is the corporeality that is dangerous: the dose makes the poison.

Similarly, as much equally water is needed to keep the states live, likewise much of it is besides risky to our health. Drinking too much water also fast can pb to a condition called water intoxication, which may be fatal. The danger in water intoxication is not that water itself becomes toxic. It is that the ingestion of likewise much water as well fast dilutes sodium ions, potassium ions, and other salts in the bloodstream to concentrations that are not high enough to back up brain, musculus, and heart functions. Military machine personnel, endurance athletes, and fifty-fifty desert hikers are susceptible to h2o intoxication if they drink water but practise not furnish the salts lost in sweat. As this example shows, even the right substances in the wrong amounts can be dangerous!

Equivalents

Concentrations of ionic solutes are occasionally expressed in units called equivalents (Eq)One mole of charge (either positive or negative). . One equivalent equals 1 mol of positive or negative charge. Thus, 1 mol/L of Na⁺(aq) is also 1 Eq/L because sodium has a 1+ charge. A ane mol/50 solution of Ca²⁺(aq) ions has a concentration of 2 Eq/L because calcium has a 2+ charge. Dilute solutions may exist expressed in milliequivalents (mEq)—for example, human blood plasma has a total concentration of about 150 mEq/L. (For more information about the ions nowadays in blood plasma, meet Chapter 3 "Ionic Bonding and Uncomplicated Ionic Compounds", Section 3.3 "Formulas for Ionic Compounds".)

Dilution

When solvent is added to dilute a solution, the volume of the solution changes, only the amount of solute does not modify. Before dilution, the amount of solute was equal to its original concentration times its original volume:

amount in moles = (concentration × volume)_initial

Subsequently dilution, the same corporeality of solute is equal to the final concentration times the last volume:

amount in moles = (concentration × volume)_final

To determine a concentration or amount later on a dilution, we can apply the following equation:

(concentration × volume)_initial = (concentration × book)_terminal

Any units of concentration and book can exist used, as long as both concentrations and both volumes take the aforementioned unit of measurement.

Example nine

A 125 mL sample of 0.900 Thou NaCl is diluted to one,125 mL. What is the last concentration of the diluted solution?

Solution

Because the volume units are the same, and we are looking for the molarity of the final solution, we can use (concentration × volume)_initial = (concentration × volume)_terminal:

(0.900 Chiliad × 125 mL) = (concentration × 1,125 mL)

We solve by isolating the unknown concentration past itself on 1 side of the equation. Dividing by one,125 mL gives

$$\text{concentration =}\frac{\text{0}\text{.900 M}\times \text{125}\cancel{\text{ mL}}}{\text{1,125}\cancel{\text{ mL}}}=0.100\text{ M}$$

as the concluding concentration.

Skill-Building Exercise

1. A nurse uses a syringe to inject 5.00 mL of 0.550 M heparin solution (heparin is an anticoagulant drug) into a 250 mL 4 bag, for a last volume of 255 mL. What is the concentration of the resulting heparin solution?

Concept Review Exercises

1. What are some of the units used to express concentration?

2. Distinguish between the terms solubility and concentration.

Answers

1. % one thousand/m, % m/v, ppm, ppb, molarity, and Eq/L (answers will vary)

2. Solubility is typically a limit to how much solute tin can dissolve in a given amount of solvent. Concentration is the quantitative amount of solute dissolved at any concentration in a solvent.

Key Takeaways

• Diverse concentration units are used to express the amounts of solute in a solution.
• Concentration units can be used as conversion factors in stoichiometry bug.
• New concentrations can be easily calculated if a solution is diluted.

Exercises

1. Define solubility. Do all solutes have the aforementioned solubility?

2. Explicate why the terms dilute or concentrated are of express usefulness in describing the concentration of solutions.

3. If the solubility of sodium chloride (NaCl) is 30.six thousand/100 mL of H₂O at a given temperature, how many grams of NaCl tin can be dissolved in 250.0 mL of H₂O?

4. If the solubility of glucose (C_viH₁₂O_six) is 120.iii g/100 mL of H_twoO at a given temperature, how many grams of C₆H₁₂O_vi tin be dissolved in 75.0 mL of H_twoO?

5. How many grams of sodium bicarbonate (NaHCO_iii) tin can a 25.0°C saturated solution have if 150.0 mL of H₂O is used as the solvent?

6. If 75.0 thou of potassium bromide (KBr) are dissolved in 125 mL of H₂O, is the solution saturated, unsaturated, or supersaturated?

7. Summate the mass/mass percentage of a saturated solution of NaCl. Use the data from Table 9.two "Solubilities of Various Solutes in Water at 25°C (Except as Noted)", presume that masses of the solute and the solvent are additive, and use the density of H₂O (1.00 g/mL) every bit a conversion factor.

8. Calculate the mass/mass percent of a saturated solution of MgCO_iii Use the information from Table 9.ii "Solubilities of Various Solutes in Water at 25°C (Except as Noted)", assume that masses of the solute and the solvent are additive, and employ the density of H₂O (1.00 g/mL) as a conversion gene.

9. Only 0.203 mL of C_sixH₆ volition dissolve in 100.000 mL of H₂O. Assuming that the volumes are additive, find the volume/book per centum of a saturated solution of benzene in water.

10. Only 35 mL of aniline (C₆H₅NH₂) volition dissolve in i,000 mL of H₂O. Assuming that the volumes are additive, discover the volume/volume percent of a saturated solution of aniline in h2o.

11. A solution of ethyl alcohol (C₂H₅OH) in water has a concentration of 20.56% v/v. What volume of C₂H_fiveOH is present in 255 mL of solution?

12. What mass of KCl is present in 475 mL of a ane.09% chiliad/five aqueous solution?

13. The average human torso contains five,830 g of blood. What mass of arsenic is present in the body if the amount in blood is 0.55 ppm?

14. The Occupational Safety and Wellness Assistants has gear up a limit of 200 ppm as the maximum safe exposure level for carbon monoxide (CO). If an boilerplate jiff has a mass of 1.286 g, what is the maximum mass of CO that can be inhaled at that maximum safe exposure level?

15. Which concentration is greater—fifteen ppm or one,500 ppb?

16. Express the concentration vii,580 ppm in parts per billion.

17. What is the molarity of 0.500 L of a potassium chromate solution containing 0.0650 mol of Thou_twoCrO_iv?

18. What is the molarity of 4.50 L of a solution containing 0.206 mol of urea [(NH₂)₂CO]?

19. What is the molarity of a 2.66 L aqueous solution containing 56.9 chiliad of NaBr?

20. If 3.08 chiliad of Ca(OH)₂ is dissolved in plenty water to make 0.875 L of solution, what is the molarity of the Ca(OH)_two?

21. What mass of HCl is present in 825 mL of a i.25 M solution?

22. What mass of isopropyl booze (C₃H₈O) is dissolved in ii.050 Fifty of a four.45 Thou aqueous C_iiiH₈O solution?

23. What volume of 0.345 M NaCl solution is needed to obtain 10.0 thousand of NaCl?

24. How many milliliters of a 0.0015 G cocaine hydrochloride (C₁₇H₂₂ClNO₄) solution is needed to obtain 0.010 g of the solute?

25. Aqueous calcium chloride reacts with aqueous silverish nitrate according to the post-obit balanced chemic equation:

    CaCl₂(aq) + 2AgNO₃(aq) → 2AgCl(s) + Ca(NO₃)₂(aq)

    How many moles of AgCl(south) are fabricated if 0.557 Fifty of 0.235 Yard CaCl₂ react with excess AgNO₃? How many grams of AgCl are made?

26. Sodium bicarbonate (NaHCO_three) is used to react with acid spills. The reaction with sulfuric acid (H₂And then₄) is equally follows:

    2NaHCO₃(southward) + H₂SO_iv(aq) → Na_twoSO₄(aq) + 2H₂O(ℓ) + 2CO₂(one thousand)

    If 27.half dozen mL of a 6.25 Grand H_iiSO₄ solution were spilled, how many moles of NaHCO₃ would be needed to react with the acid? How many grams of NaHCO₃ is this?

27. The fermentation of glucose to make ethanol and carbon dioxide has the following overall chemical equation:

    C_viH₁₂O₆(aq) → 2C₂H_vOH(aq) + 2CO_ii(m)

    If 1.00 50 of a 0.567 M solution of C₆H₁₂O₆ were completely fermented, what would be the resulting concentration of the C_twoH₅OH solution? How many moles of CO_ii would exist formed? How many grams is this? If each mole of CO₂ had a volume of 24.5 L, what volume of CO₂ is produced?

28. Aqueous sodium bisulfite gives off sulfur dioxide gas when heated:

    2NaHSO₃(aq) → Na_iiSO_iii(aq) + H_iiO(ℓ) + SO₂(g)

    If 567 mL of a i.005 M NaHSO₃ solution were heated until all the NaHSO₃ had reacted, what would exist the resulting concentration of the Na_iiSO₃ solution? How many moles of SO_ii would be formed? How many grams of SO₂ would be formed? If each mole of Then₂ had a volume of 25.78 Fifty, what volume of So₂ would be produced?

29. What is the concentration of a i.0 M solution of K⁺(aq) ions in equivalents/liter?

30. What is the concentration of a ane.0 M solution of And so_four ²⁻(aq) ions in equivalents/liter?

31. A solution having initial concentration of 0.445 M and initial volume of 45.0 mL is diluted to 100.0 mL. What is its final concentration?

32. A 50.0 mL sample of saltwater that is three.0% m/v is diluted to 950 mL. What is its final mass/volume percent?

Answers

1. Solubility is the amount of a solute that can dissolve in a given amount of solute, typically 100 mL. The solubility of solutes varies widely.

3. 76.5 yard

5. 12.six g

7. 26.5%

9. 0.203%

11. 52.iv mL

13. 0.00321 1000

15. 15 ppm

17. 0.130 G

19. 0.208 M

21. 37.six g

23. 0.496 L

25. 0.262 mol; 37.five g

27. i.xiii K C₂H₅OH; 1.xiii mol of CO₂; 49.vii yard of CO_ii; 27.7 L of CO₂

29. 1.0 Eq/L

31. 0.200 M

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Source: http://saylordotorg.github.io/text_the-basics-of-general-organic-and-biological-chemistry/s12-02-concentration.html