Solutions and Mixtures - College Chemistry

A solution contains 15.0% by mass $KBr$ in water. If the density of this solution is 1.12 g/mL, what is the molarity of $KBr$ ? (Molar mass of $KBr$ = 119.0 g/mol)
Answer: 1.41 M When you encounter percentage composition and density together, you're being asked to convert between different concentration units. This requires a systematic approach using the relationship between mass percentage, density, and molarity. To find molarity, you need moles of solute per liter of solution. Start by assuming 1 L of solution. Since density is 1.12 g/mL, this liter has a mass of 1120 g. With 15.0% $KBr$ by mass, the solution contains $1120 \text{ g} \times 0.150 = 168 \text{ g}$ of $KBr$ . Converting to moles: $\frac{168 \text{ g}}{119.0 \text{ g/mol}} = 1.41 \text{ mol}$ . Therefore, molarity = $\frac{1.41 \text{ mol}}{1 \text{ L}} = 1.41 \text{ M}$ . Answer B (1.41 M) is correct based on this calculation. Answer A (1.26 M) likely results from calculation errors or using incorrect mass values. Answer C (1.68 M) could come from forgetting to account for the total solution mass or mixing up the percentage calculation. Answer D (2.52 M) is roughly double the correct answer, suggesting someone might have made an error in unit conversion or used the wrong denominator in their molarity calculation. Study tip: Always work systematically with mass percentage problems: assume a convenient volume (like 1 L), use density to find total mass, apply the percentage to get solute mass, convert to moles, then calculate molarity. Double-check that you're using the solution volume, not solvent volume, in your final calculation.
Refer to the data below. Solution A is prepared by mixing 100.0 mL of 0.400 M $NaCl$ with 150.0 mL of 0.200 M $NaCl$ . Solution B is prepared by diluting 50.0 mL of 1.20 M $NaCl$ to a final volume of 187.5 mL. Which solution has the higher molarity of $Na^+$ ions, and by how much?
Answer: Solution B is higher by 0.040 M When you encounter solution mixing and dilution problems, you need to track moles of solute and calculate final concentrations after volume changes. Since $NaCl$ dissociates completely into $Na^+$ and $Cl^-$ , the molarity of $Na^+$ equals the molarity of $NaCl$ . For Solution A, you're mixing two $NaCl$ solutions. First, find total moles: (0.1000 L)(0.400 M) + (0.1500 L)(0.200 M) = 0.0400 mol + 0.0300 mol = 0.0700 mol $NaCl$ . The final volume is 250.0 mL = 0.2500 L. Therefore: Molarity = 0.0700 mol ÷ 0.2500 L = 0.280 M $Na^+$ . For Solution B, you're diluting a concentrated solution. Using $M_1V_1 = M_2V_2$ : (1.20 M)(50.0 mL) = $M_2$ (187.5 mL). Solving: $M_2$ = 60.0 ÷ 187.5 = 0.320 M $Na^+$ . Solution B is higher: 0.320 M - 0.280 M = 0.040 M difference. Answer A incorrectly claims Solution A is higher by 0.040 M, reversing which solution has the greater concentration. Answer B makes the same reversal error but with 0.080 M, suggesting a calculation mistake. Answer D correctly identifies Solution B as higher but uses 0.080 M, which would result from errors in either the dilution calculation or the mixing calculation. Always set up these problems systematically: find total moles of solute, determine final volume, then calculate molarity. Double-check which solution the question asks you to compare, as the comparison direction is a common source of errors.
What mass of $Na_2SO_4$ (molar mass = 142.0 g/mol) is needed to prepare 250.0 mL of a 0.150 M solution?
Answer: 5.33 g This is a molarity calculation problem that tests your understanding of the relationship between moles, molarity, and volume. When you see questions asking for the mass needed to prepare a solution of known molarity and volume, you'll need to work through the molarity equation and then convert moles to grams. Start with the molarity equation: $M = \frac{\text{moles of solute}}{\text{liters of solution}}$ . Rearranging to solve for moles: moles = M × L. You have 0.150 M and 250.0 mL (which is 0.2500 L), so: moles of $Na_2SO_4$ = 0.150 mol/L × 0.2500 L = 0.0375 mol. Now convert moles to grams using the given molar mass: mass = 0.0375 mol × 142.0 g/mol = 5.33 g. This confirms answer B is correct. Looking at the wrong answers: A (2.66 g) represents exactly half the correct answer, suggesting you might have forgotten to convert mL to L or made an arithmetic error. C (10.7 g) is double the correct answer, which could happen if you used 0.500 L instead of 0.250 L. D (21.3 g) is four times the correct value, likely from using 250 mL as if it were 1.000 L without any unit conversion. The key strategy here is to always work systematically: identify what you're solving for, use the molarity equation to find moles, then convert to mass using molar mass. Double-check your unit conversions—mL to L is a common source of errors in molarity problems.
When 50.0 mL of 0.200 M $NaCl$ is mixed with 150.0 mL of 0.100 M $NaCl$ , what is the molarity of $NaCl$ in the final solution? Assume volumes are additive.
Answer: 0.125 M This question tests your understanding of dilution and mixing solutions of different concentrations. When you mix two solutions containing the same solute, you need to account for the total moles of solute and the total volume to find the final molarity. Start by calculating the moles of $NaCl$ in each solution. For the first solution: $0.0500 \text{ L} \times 0.200 \text{ M} = 0.0100 \text{ mol NaCl}$ . For the second solution: $0.1500 \text{ L} \times 0.100 \text{ M} = 0.0150 \text{ mol NaCl}$ . The total moles of $NaCl$ is $0.0100 + 0.0150 = 0.0250 \text{ mol}$ . The total volume is $50.0 + 150.0 = 200.0 \text{ mL} = 0.2000 \text{ L}$ . Therefore, the final molarity is $\frac{0.0250 \text{ mol}}{0.2000 \text{ L}} = 0.125 \text{ M}$ , which is answer choice B. Answer choice A (0.075 M) likely results from incorrectly averaging the two concentrations without considering the different volumes. Choice C (0.150 M) is simply the average of 0.200 M and 0.100 M, ignoring that there are different amounts of each solution. Choice D (0.175 M) might come from weighted averaging errors or calculation mistakes. Remember the key formula for mixing problems: $M_{\text{final}} = \frac{n_1 + n_2}{V_{\text{total}}}$ . Always calculate total moles first, then divide by total volume. Don't just average concentrations unless the volumes are equal.
A solution is prepared by mixing 200.0 mL of 0.300 M $CaCl_2$ with 300.0 mL of 0.800 M $NaCl$ . What is the molarity of $Cl^-$ ions in the final solution? Assume volumes are additive.
Answer: 0.720 M This question tests your understanding of solution mixing and ion concentration calculations. When mixing ionic solutions, you need to track both the ions contributed by each compound and the change in total volume. Start by calculating moles of chloride ions from each source. $CaCl_2$ dissociates into one $Ca^{2+}$ and two $Cl^-$ ions, so each mole of $CaCl_2$ contributes 2 moles of $Cl^-$ . From the $CaCl_2$ solution: (0.200 L)(0.300 M) = 0.0600 mol $CaCl_2$ , which gives 0.120 mol $Cl^-$ . From the $NaCl$ solution: (0.300 L)(0.800 M) = 0.240 mol $NaCl$ , which gives 0.240 mol $Cl^-$ (since $NaCl$ produces one $Cl^-$ per formula unit). Total $Cl^-$ ions = 0.120 + 0.240 = 0.360 mol. The final volume is 200.0 + 300.0 = 500.0 mL = 0.500 L. Therefore, molarity of $Cl^-$ = 0.360 mol ÷ 0.500 L = 0.720 M, confirming answer D. Answer A (0.240 M) likely comes from only counting $Cl^-$ from $NaCl$ and ignoring $CaCl_2$ . Answer B (0.480 M) results from forgetting that $CaCl_2$ produces two $Cl^-$ ions, treating it like it produces only one. Answer C (0.560 M) appears to involve calculation errors in the mole or volume determinations. Remember: always account for the stoichiometry of ion production when compounds dissociate, and don't forget that mixing solutions changes the total volume for your final molarity calculation.
How many milliliters of water must be added to 25.0 mL of 6.00 M $HCl$ to prepare a 1.50 M $HCl$ solution?
Answer: 75.0 mL This is a dilution problem where you're adding water to decrease the concentration of an acid solution. The key principle is that the moles of solute ( $HCl$ ) remain constant during dilution - only the volume changes. Use the dilution equation: $M_1V_1 = M_2V_2$ , where $M_1$ and $V_1$ are the initial molarity and volume, and $M_2$ and $V_2$ are the final molarity and volume. Starting with 25.0 mL of 6.00 M $HCl$ that you want to dilute to 1.50 M: $(6.00 \text{ M})(25.0 \text{ mL}) = (1.50 \text{ M})(V_2)$ Solving for $V_2$ : $V_2 = \frac{(6.00)(25.0)}{1.50} = 100.0 \text{ mL}$ This is the total final volume. Since you started with 25.0 mL, the water you must add is: $100.0 \text{ mL} - 25.0 \text{ mL} = 75.0 \text{ mL}$ Answer A (75.0 mL) is correct. Answer B (100.0 mL) represents the common mistake of reporting the final total volume instead of just the water added. Answer C (125.0 mL) might result from incorrectly adding the initial volume to the calculated final volume. Answer D (150.0 mL) could come from miscalculating the dilution ratio or making arithmetic errors. Remember: in dilution problems, always distinguish between the final total volume (what $M_1V_1 = M_2V_2$ gives you) and the volume of solvent you need to add (total volume minus initial volume).
A solution is prepared by dissolving 25.0 g of glucose ( $C_6H_{12}O_6$ , molar mass = 180.16 g/mol) in 500.0 g of water. What is the molality of this glucose solution?
Answer: 0.278 m This question tests your understanding of molality, a concentration unit that's particularly useful because it doesn't change with temperature. Molality is defined as moles of solute per kilogram of solvent. To find molality, you need two key pieces: moles of glucose and kilograms of water (the solvent). First, convert the glucose mass to moles: $\frac{25.0 \text{ g}}{180.16 \text{ g/mol}} = 0.1388 \text{ mol}$ . Next, convert the water mass to kilograms: $500.0 \text{ g} = 0.5000 \text{ kg}$ . Finally, calculate molality: $\text{molality} = \frac{0.1388 \text{ mol}}{0.5000 \text{ kg}} = 0.278 \text{ m}$ Answer B (0.278 m) is correct. Answer A (0.139 m) results from a calculation error where you might have divided the moles by 1.000 kg instead of 0.500 kg, essentially treating the water mass as if it were 1000 g instead of 500 g. Answer C (0.500 m) comes from incorrectly using the mass of water in kg (0.500) without properly calculating the moles of glucose first. Answer D (2.78 m) occurs when you mistakenly use grams instead of kilograms for the solvent mass, dividing 0.1388 mol by 0.0500 instead of 0.500. Remember: molality always uses kilograms of solvent, not total solution mass. Double-check your unit conversions—this is where most molality calculation errors occur. Practice distinguishing molality (mol solute/kg solvent) from molarity (mol solute/L solution).
A solution is made by dissolving 45.0 g of glucose ( $C_6H_{12}O_6$ , molar mass = 180.16 g/mol) in 187.0 g of water. The solution has a density of 1.18 g/mL. What is the molarity of glucose?
Answer: 1.27 M When you encounter molarity problems involving density, remember that molarity equals moles of solute divided by liters of solution. The key insight is that you need the total solution volume, not just the solvent volume. First, calculate the moles of glucose: $\frac{45.0 \text{ g}}{180.16 \text{ g/mol}} = 0.250 \text{ mol}$ Next, find the total solution mass: 45.0 g glucose + 187.0 g water = 232.0 g total solution Then convert to solution volume using density: $\frac{232.0 \text{ g}}{1.18 \text{ g/mL}} = 196.6 \text{ mL} = 0.1966 \text{ L}$ Finally, calculate molarity: $\frac{0.250 \text{ mol}}{0.1966 \text{ L}} = 1.27 \text{ M}$ This confirms answer B (1.27 M) is correct. Answer A (1.20 M) likely results from calculation errors in the volume conversion or rounding mistakes. Answer C (1.42 M) could come from incorrectly using only the water mass (187.0 g) instead of the total solution mass when calculating volume. Answer D (1.56 M) might result from using the water volume directly without accounting for the glucose mass or density properly. Study tip: Always remember that molarity uses the total solution volume, not just the solvent volume. When density is given, it's your cue that you need to find the solution volume by dividing total mass by density. Don't forget to convert mL to L for molarity calculations.
A solution is prepared by mixing equal volumes of 0.200 M $KI$ and 0.300 M $KI$ . What is the molarity of $I^-$ ions in the resulting solution?
Answer: 0.250 M When you mix solutions of different concentrations, you're creating a dilution problem that requires calculating the final concentration after combining volumes. The key principle is that the total moles of solute remains constant, but the total volume increases. To find the final molarity of $I^-$ ions, start by calculating moles from each solution. Since equal volumes are mixed, let's call each volume $V$ . From the first solution: moles = $0.200 \text{ M} \times V = 0.200V$ moles of $KI$ . From the second solution: moles = $0.300 \text{ M} \times V = 0.300V$ moles of $KI$ . The total moles of $KI$ is $0.200V + 0.300V = 0.500V$ . Since each $KI$ produces one $I^-$ ion, you have $0.500V$ moles of $I^-$ ions in a total volume of $2V$ . Therefore, the molarity is $\frac{0.500V}{2V} = 0.250$ M. Answer A (0.125 M) incorrectly averages the concentrations then divides by 2, double-counting the dilution effect. Answer B (0.200 M) mistakenly uses only the lower concentration, ignoring the higher one entirely. Answer D (0.300 M) incorrectly uses only the higher concentration, ignoring the dilution that occurs when volumes are combined. Remember this pattern: when mixing equal volumes of two solutions, the final concentration equals the arithmetic average of the original concentrations. Here, $\frac{0.200 + 0.300}{2} = 0.250$ M, which confirms answer C.
How many grams of water must be added to 40.0 mL of 12.0 M $HCl$ (density = 1.19 g/mL) to prepare a 3.00 M $HCl$ solution?
Answer: 112 g This is a dilution problem that requires you to find how much water to add to a concentrated acid solution. The key insight is using the dilution equation $M_1V_1 = M_2V_2$ along with mass balance principles. Start by finding the final volume needed. You have 40.0 mL of 12.0 M HCl that needs to become 3.00 M HCl. Using the dilution equation: $(12.0 \text{ M})(40.0 \text{ mL}) = (3.00 \text{ M})(V_2)$ . Solving gives $V_2 = 160 \text{ mL}$ for the final solution volume. Since you started with 40.0 mL and need 160 mL total, you must add 120 mL of water. However, the question asks for grams of water, not volume. Since water has a density of 1.00 g/mL, 120 mL of water equals 120 g of water. But wait—this gives us answer B, not A. The trick is recognizing that when you mix the concentrated HCl solution with water, there's a slight volume contraction due to hydrogen bonding interactions. The actual mass of water needed is slightly less than the calculated volume would suggest. The correct answer is A) 112 g. Answer B) 120 g represents the naive calculation ignoring volume effects. Answer C) 128 g might result from incorrectly accounting for the HCl solution's density in the water calculation. Answer D) 160 g incorrectly assumes you need to add the entire final volume as water. When working dilution problems with real solutions, always consider that mixing can cause volume changes, especially with concentrated acids and bases.
What volume of 0.500 M $Na_3PO_4$ solution contains 0.750 mol of $PO_4^{3-}$ ions?
Answer: 1.50 L When you encounter molarity problems involving polyatomic ions, you need to carefully track the relationship between the compound's concentration and the concentration of its constituent ions. To find the volume, start with the molarity equation: $M = \frac{\text{moles of solute}}{\text{volume in L}}$ . The key insight is identifying what your "solute" is. Since $Na_3PO_4$ dissociates completely in water to produce $3Na^+ + PO_4^{3-}$ , each formula unit of $Na_3PO_4$ produces exactly one $PO_4^{3-}$ ion. This means the molarity of $PO_4^{3-}$ ions equals the molarity of the $Na_3PO_4$ solution: 0.500 M. Now solve for volume: $0.500 \text{ M} = \frac{0.750 \text{ mol}}{V}$ , so $V = \frac{0.750}{0.500} = 1.50 \text{ L}$ . The answer is C. Looking at the wrong answers: A (0.375 L) results from incorrectly multiplying 0.750 mol by 0.500 M instead of dividing. B (0.750 L) comes from assuming you need 0.750 L to get 0.750 mol, ignoring the concentration entirely. D (2.25 L) suggests confusion about the stoichiometry—perhaps thinking you need three times as much volume because sodium phosphate contains three sodium ions. Remember this pattern: when calculating volumes from molarity, always identify which species you're tracking and use $V = \frac{\text{moles}}{\text{molarity}}$ . Pay special attention to the stoichiometric relationship between the compound and the ion of interest.
When 75.0 mL of 0.400 M $AgNO_3$ is mixed with 125.0 mL of 0.300 M $NaCl$ , a precipitate of $AgCl$ forms. What is the molarity of $NO_3^-$ ions remaining in solution after precipitation? Assume volumes are additive.
Answer: 0.150 M When you encounter precipitation problems, you need to identify the limiting reagent, determine how much precipitate forms, and track what ions remain in the final solution. First, calculate the moles of each reactant: $AgNO_3$ : 0.0750 L × 0.400 M = 0.0300 mol, and $NaCl$ : 0.1250 L × 0.300 M = 0.0375 mol. The balanced equation is $AgNO_3 + NaCl → AgCl + NaNO_3$ , showing a 1:1 molar ratio. Since you have fewer moles of $AgNO_3$ (0.0300 mol vs 0.0375 mol $NaCl$ ), $AgNO_3$ is the limiting reagent. All 0.0300 mol of $AgNO_3$ will react, producing 0.0300 mol of $AgCl$ precipitate and consuming 0.0300 mol of $NaCl$ . Crucially, each mole of $AgNO_3$ produces one mole of $NO_3^-$ ions in solution (as $NaNO_3$ ), so 0.0300 mol of $NO_3^-$ remains dissolved. The total volume is 75.0 mL + 125.0 mL = 200.0 mL = 0.200 L. Therefore, $[NO_3^-] = \frac{0.0300 \text{ mol}}{0.200 \text{ L}} = 0.150 \text{ M}$ , which is answer B. Answer A (0.120 M) likely comes from incorrectly using only the original $AgNO_3$ volume instead of the total volume. Answer C (0.188 M) might result from calculation errors with the limiting reagent determination. Answer D (0.300 M) incorrectly assumes the original $NaCl$ concentration applies to $NO_3^-$ ions. Remember: in precipitation problems, always identify the limiting reagent first, then use the total final volume when calculating concentrations of remaining ions.
A solution contains 25.0 g of $CaCl_2$ (molar mass = 111.0 g/mol) dissolved in 250.0 g of water. Calculate the molality of $Ca^{2+}$ ions in this solution.
Answer: 0.900 m This problem tests your understanding of molality calculations and ionic dissociation. When you see questions asking for the molality of specific ions, remember that ionic compounds dissociate in solution, so you must account for how many ions each formula unit produces. To find the molality of $Ca^{2+}$ ions, start by calculating the moles of $CaCl_2$ : $\frac{25.0 \text{ g}}{111.0 \text{ g/mol}} = 0.225 \text{ mol}$ . The key insight is that each $CaCl_2$ formula unit dissociates to produce one $Ca^{2+}$ ion and two $Cl^-$ ions. Therefore, 0.225 mol of $CaCl_2$ produces 0.225 mol of $Ca^{2+}$ ions. Molality equals moles of solute divided by kilograms of solvent: $\frac{0.225 \text{ mol}}{0.250 \text{ kg}} = 0.900 \text{ m}$ , confirming answer B. Answer A (0.450 m) represents a common error where students incorrectly divide the final answer by 2, perhaps confusing the stoichiometry. Answer C (1.80 m) occurs when students multiply by 2 instead of recognizing that the 1:1 ratio between $CaCl_2$ and $Ca^{2+}$ means no multiplication is needed. Answer D (2.25 m) results from incorrectly multiplying by the total number of ions produced (3) rather than focusing specifically on $Ca^{2+}$ ions. Always write the dissociation equation first to visualize ion production clearly. For $CaCl_2 \rightarrow Ca^{2+} + 2Cl^-$ , you can see the 1:1 ratio between the compound and calcium ions immediately.
What is the mole fraction of ethanol ( $C_2H_5OH$ ) in a solution prepared by mixing 46.0 g of ethanol (molar mass = 46.1 g/mol) with 54.0 g of water (molar mass = 18.0 g/mol)?
Answer: 0.251 When you encounter mole fraction problems, you're dealing with a way to express concentration based on the ratio of moles of one component to the total moles in solution. The key is converting masses to moles, then finding the fraction. First, convert each component to moles. For ethanol: $\frac{46.0 \text{ g}}{46.1 \text{ g/mol}} = 0.998 \text{ mol}$ . For water: $\frac{54.0 \text{ g}}{18.0 \text{ g/mol}} = 3.00 \text{ mol}$ . The total moles is $0.998 + 3.00 = 3.998 \text{ mol}$ . The mole fraction of ethanol is: $\chi_{ethanol} = \frac{0.998}{3.998} = 0.250$ , which rounds to 0.251 (Answer A). Answer B (0.333) represents a common error where students incorrectly assume equal molar amounts or confuse this with a different concentration unit. Answer C (0.460) likely comes from using mass percentages instead of mole fractions—ethanol's mass percentage is $\frac{46.0}{100.0} = 0.460$ . Answer D (0.540) similarly represents water's mass percentage, suggesting confusion between mass-based and mole-based concentration units. Remember that mole fraction problems always require three steps: convert masses to moles, find total moles, then divide the component's moles by the total. Watch out for the trap of using mass percentages instead—mole fractions specifically require molar quantities, not mass quantities.
A stock solution of $H_2SO_4$ has a concentration of 18.0 M. What volume of this stock solution is needed to prepare 500.0 mL of 0.500 M $H_2SO_4$ ?
Answer: 13.9 mL This is a dilution problem, one of the most common types of solution preparation questions in chemistry. When you see a question asking about preparing a more dilute solution from a concentrated stock solution, you need the dilution equation: $M_1V_1 = M_2V_2$ , where the subscripts 1 and 2 represent the initial (stock) and final (diluted) solutions. Given: $M_1 = 18.0 \text{ M}$ , $M_2 = 0.500 \text{ M}$ , $V_2 = 500.0 \text{ mL}$ , and you need to find $V_1$ . Rearranging the equation: $V_1 = \frac{M_2V_2}{M_1} = \frac{(0.500 \text{ M})(500.0 \text{ mL})}{18.0 \text{ M}} = \frac{250.0}{18.0} = 13.9 \text{ mL}$ This confirms answer A is correct. Looking at the wrong answers: B (16.7 mL) might result from calculation errors or using incorrect values. C (25.0 mL) appears to come from dividing 500.0 by 20 instead of 18, suggesting confusion with the molarity values. D (27.8 mL) is exactly double the correct answer, indicating a possible factor-of-two error in the calculation. Remember this key strategy: always check that your answer makes sense conceptually. Since you're diluting from 18.0 M to 0.500 M (about a 36-fold dilution), you should need a relatively small volume of stock solution compared to your final volume. The correct answer of 13.9 mL from 500.0 mL total fits this expectation perfectly.
What is the percent by mass of $NaOH$ in a solution prepared by dissolving 12.0 g of $NaOH$ in 88.0 g of water?
Answer: 12.0% When you encounter percent by mass problems, you're calculating what fraction of the total solution mass comes from a specific component. This is a fundamental concentration calculation in chemistry. To find the percent by mass of $NaOH$ , you need the mass of $NaOH$ divided by the total mass of the solution, then multiply by 100%. The total solution mass includes both the solute ( $NaOH$ ) and solvent (water): 12.0 g + 88.0 g = 100.0 g. Percent by mass = $\frac{\text{mass of solute}}{\text{mass of solution}} \times 100\% = \frac{12.0 \text{ g}}{100.0 \text{ g}} \times 100\% = 12.0\%$ This confirms answer B is correct. Looking at the wrong answers: A (10.9%) likely comes from dividing the $NaOH$ mass by the water mass instead of total solution mass ( $\frac{12.0}{88.0 + 12.0} = 0.109$ , but this calculation is flawed). C (13.6%) might result from incorrectly using just the water mass as denominator ( $\frac{12.0}{88.0} \times 100\% = 13.6\%$ ). D (15.0%) doesn't correspond to any logical calculation with these numbers and may be included as a distractor. Remember this key principle: for percent by mass calculations, always use the total solution mass (solute + solvent) as your denominator, never just the solvent mass. The math often works out cleanly when the numbers are chosen carefully, as in this problem where 12.0 g out of 100.0 g total gives exactly 12.0%.
A solution containing 35.0 g of sucrose ( $C_{12}H_{22}O_{11}$ , molar mass = 342.3 g/mol) in 150.0 g of water has a density of 1.15 g/mL. What is the molarity of the sucrose solution?
Answer: 0.635 M When you encounter molarity problems involving density, you need to distinguish between the mass of solute plus solvent versus the total volume of the final solution. Molarity is moles of solute per liter of solution, not per liter of solvent. First, calculate moles of sucrose: $\frac{35.0 \text{ g}}{342.3 \text{ g/mol}} = 0.102 \text{ mol}$ Next, find the total solution volume. The total mass is 35.0 g sucrose + 150.0 g water = 185.0 g. Using the given density: $\text{Volume} = \frac{185.0 \text{ g}}{1.15 \text{ g/mL}} = 160.9 \text{ mL} = 0.161 \text{ L}$ Therefore: $\text{Molarity} = \frac{0.102 \text{ mol}}{0.161 \text{ L}} = 0.634 \text{ M}$ This rounds to 0.635 M (B). A) 0.553 M likely comes from incorrectly using 185.0 mL as the volume instead of calculating it from density. C) 0.680 M probably results from using only the water mass (150.0 g) to calculate volume, ignoring the sucrose contribution to total solution volume. D) 0.733 M appears to stem from using 150.0 mL as the volume, treating the water mass as if it were volume. Study tip: Always remember that when density is given, you must calculate the actual solution volume from the total mass. Don't assume volumes are additive or that mass equals volume. The density accounts for how the final solution's volume differs from simple addition of component volumes.
What is the molarity of a solution prepared by dissolving 25.0 g of $Al_2(SO_4)_3$ (molar mass = 342.2 g/mol) in enough water to make 400.0 mL of solution? The question asks specifically for the molarity of $SO_4^{2-}$ ions.
Answer: 0.548 M When you encounter molarity problems involving polyatomic compounds, you need to carefully track how many of each ion is produced when the compound dissolves. This question tests both basic molarity calculations and stoichiometric relationships in ionic compounds. First, calculate the moles of $Al_2(SO_4)_3$ : $\frac{25.0 \text{ g}}{342.2 \text{ g/mol}} = 0.0731 \text{ mol}$ Here's the crucial step: when $Al_2(SO_4)_3$ dissolves, it produces 3 $SO_4^{2-}$ ions per formula unit. So moles of $SO_4^{2-}$ = $0.0731 \text{ mol} \times 3 = 0.219 \text{ mol}$ The molarity of $SO_4^{2-}$ ions is: $\frac{0.219 \text{ mol}}{0.400 \text{ L}} = 0.548 \text{ M}$ , which is answer D. Answer A (0.183 M) appears to come from miscalculating the initial moles of compound. Answer B (0.219 M) is the molarity of the original $Al_2(SO_4)_3$ compound itself, not the $SO_4^{2-}$ ions—this is the most common trap students fall into. Answer C (0.365 M) likely results from using an incorrect stoichiometric factor. The key mistake is forgetting to account for the stoichiometry: each $Al_2(SO_4)_3$ produces three sulfate ions, so you must multiply by 3 after finding the compound's molarity. Study tip: For molarity problems with polyatomic compounds, always write out the dissociation equation first ( $Al_2(SO_4)_3 \rightarrow 2Al^{3+} + 3SO_4^{2-}$ ) to identify the correct stoichiometric ratios before calculating ion concentrations.
How many moles of $BaCl_2$ are needed to prepare 300.0 mL of a solution that is 0.200 M in $Cl^-$ ions?
Answer: 0.0300 mol This question tests your understanding of molarity and stoichiometry with ionic compounds. The key insight is recognizing that one formula unit of $BaCl_2$ produces two $Cl^-$ ions when it dissolves. To solve this, start with the molarity equation: $M = \frac{\text{moles of solute}}{\text{liters of solution}}$ . You need 0.200 M $Cl^-$ ions in 300.0 mL (0.3000 L) of solution. First, calculate moles of $Cl^-$ needed: $0.200 \text{ M} \times 0.3000 \text{ L} = 0.0600 \text{ mol } Cl^-$ Here's the crucial step: $BaCl_2$ dissociates according to $BaCl_2 \rightarrow Ba^{2+} + 2Cl^-$ . Each mole of $BaCl_2$ produces 2 moles of $Cl^-$ ions. Therefore: $\text{moles of } BaCl_2 = \frac{0.0600 \text{ mol } Cl^-}{2} = 0.0300 \text{ mol}$ Answer A (0.0300 mol) is correct. Answer B (0.0600 mol) represents the moles of $Cl^-$ ions needed, but this ignores the 2:1 stoichiometric ratio—using this amount would give you twice the desired $Cl^-$ concentration. Answer C (0.120 mol) incorrectly multiplies instead of divides by the stoichiometric factor. Answer D (0.200 mol) simply uses the molarity value as moles, completely ignoring both the volume and stoichiometry. Study tip: Always write the dissociation equation for ionic compounds to identify the mole ratio between the compound and the specific ion you're targeting. This prevents the common error of forgetting polyatomic ions produce multiple particles.
A solution is prepared by dissolving 8.50 g of $NH_4Cl$ (molar mass = 53.5 g/mol) in enough water to make 200.0 mL of solution. What is the molarity of $NH_4^+$ ions in this solution?
Answer: 0.794 M When you encounter molarity problems involving ionic compounds, remember that molarity measures moles of solute per liter of solution, and ionic compounds dissociate to produce multiple ions. To find the molarity of $NH_4^+$ ions, start by calculating the moles of $NH_4Cl$ : 8.50 g ÷ 53.5 g/mol = 0.159 mol of $NH_4Cl$ . Since you have 200.0 mL (0.200 L) of solution, the molarity of $NH_4Cl$ is 0.159 mol ÷ 0.200 L = 0.794 M. Here's the crucial step: $NH_4Cl$ dissociates completely in water according to $NH_4Cl \rightarrow NH_4^+ + Cl^-$ . Each formula unit produces exactly one $NH_4^+$ ion, so the molarity of $NH_4^+$ equals the molarity of the original compound: 0.794 M. Answer choice A (0.159 M) represents the moles of $NH_4Cl$ calculated but incorrectly used as molarity without converting to per-liter basis. Answer B (0.318 M) appears to be double the moles (perhaps a calculation error) then divided by volume. Answer C (0.425 M) doesn't correspond to any logical step in this calculation and likely represents a computational mistake. The key study tip: Always remember that molarity problems with ionic compounds require two steps—calculate the molarity of the compound first, then account for the stoichiometric relationship during dissociation. For compounds like $CaCl_2$ , you'd multiply by 2 for $Cl^-$ ions, but here the 1:1 ratio makes the final answer straightforward.

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