Chem Fall 2018 -Solutions
While not used as frequently, mole fraction is another unit of concentration. Mole fraction is the fraction of moles of one substance over the total moles in the sample. mole fraction (chi) equals moles of A over total moles What is the mole fraction of O2 in a sample with 3.50 grams of O2 and 25.6 g N2?
0.107 The mole fraction of O2 in a sample with 3.50 grams of O2 and 25.6 grams of N2 is 0.107. Note that mole fractions are reported without units. Find the moles of each substance using molar mass. 3.50 grams O2 times 1 mole O2 over 32.00 grams O2 equals 0.109 moles O2 25.6 grams N2 times 1 mole N2 over 28.02 grams N2 equals 0.914 moles N2 Then, calculate the mole fraction of O2. chi equals 0.109 moles O2 over the sum of 0.109 moles O2 and 0.914 moles N2 so chi equals 0.107
Molarity is one of the most commonly used concentration units for solutions and is abbreviated as "M." Molarity is calculated by determining the number of moles of the solute, or the substance being dissolved, in 1 liter of solution. In other words, molarity is the number of moles/liter. This is summarized as follows: molarity equals moles of solute over 1 liter of solution What is the molarity of a solution containing 45.3 grams of Mg(NO3)2 and a volume of 2.25 L?
0.136 M Mg(NO3)2 The molarity of a solution containing 45.3 grams of Mg(NO3)2 and a volume of 2.25 L is 0.136 M Mg(NO3)2. Since molarity is the number of moles in 1 liter, the number of grams must be converted from grams (g) to moles using the molar mass. 45.3 grams magnesium nitrate over one times 1 mole magnesium nitrate over 148.33 grams magnesium nitrate equals 0.305 moles magnesium nitrate The molarity would then be calculated according to the following equation: molarity equals 0.305 moles magnesium nitrate over 2.25 liters equals 0.136 moles per liter magnesium nitrate or 0.136 molar magnesium nitrate
While not used as frequently, mole fraction is another unit of concentration. Mole fraction is the fraction of moles of one substance over the total moles in the sample. mole fraction (chi) equals moles of A over total moles What is the mole fraction of O2 in a sample with 1.25 grams CH4, 45.2 g N2, and 13.4 g O2?
0.199 The mole fraction of O2 in a sample with 1.25 grams of CH4, 45.2 grams of N2, and 13.4 grams of O2 is 0.199. Note that mole fractions are reported without units. Find the moles of each substance using molar mass. 1.25 grams CH4 times 1 mole CH4 over 16.05 grams CH4 equals 0.0779 moles CH4 45.2 grams N2 times 1 mole N2 over 28.02 grams N2 equals 1.61 moles N2 13.4 grams O2 times 1 mole O2 over 32.00 grams O2 equals 0.419 moles O2 Then, calculate the mole fraction of O2. chi equals 0.419 moles O2 over the sum of 0.0779 moles CH4, 1.61 moles N2, and 0.419 moles O2 so chi equals 0.199
Which substance is nonpolar?
C4H10 (butane)
Which substance is nonpolar?
CO2 The nonpolar substance is CO2 because it has no net dipole moment. The remaining substances (HCl, H2S, and KBr) are all polar. Polarity results when a molecule has a net dipole moment resulting from the dipole moments of individual bonds not cancelling out. The dipole moment of the bond results from an uneven sharing of electrons.
Colligative properties are those that depend on the number of particles dissolved in a solvent. The more particles in the solvent, the greater the effect the solute has on the properties of the solution. Examples of colligative properties include boiling point elevation, freezing point depression, vapor pressure depression, and osmotic pressure. What is the freezing point of an aqueous solution with an NaCl concentration of 1.85 m? Kf,water = 1.853°C/m
−6.86°C
The ability for a substance to dissolve is affected by both temperature and pressure. Temperature As the temperature increases, the solubility of a solid solute also increases, as the solvent can hold more of the solute particles. However, as the temperature increases for a gas, the solubility decreases. As the temperature increases, the gas molecules have more energy and are able to escape the solvent, causing the solubility to decrease. Pressure As the pressure increases on a solution containing a gaseous solute, the solubility of the gas increases. At higher pressures, there are more gas molecules above the solution, which are able to enter and dissolve into the solution. How much KNO3 can dissolve in 50 g of water at 10°C? Graph representing how temperature affects the solubility of various solids. At 40°C, Glucose has a solubility of ~130 g in 100 g of water while NaCl has a solubility of ~38 g in 100 g of water.
10 g At 10°C, 10 g of KNO3 will dissolve in 50 g of water. Based on the solubility curve, 20 g of KNO3 can dissolve in 100 g of water at 10°C. Since the mass of water is halved, the amount of KNO3 must also be halved.
Colligative properties are those that depend on the number of particles dissolved in a solvent. The more particles in the solvent, the greater the effect the solute has on the properties of the solution. Examples of colligative properties include boiling point elevation, freezing point depression, vapor pressure depression, and osmotic pressure. Which aqueous solution has the lowest vapor pressure?
χglucose = 0.25 The solution with the lowest vapor pressure is glucose with χ = 0.25. The equation for vapor pressure of a solution is P°=χsolvent∙P where P° is the vapor pressure of the solution, χ is the mole fraction of the solvent, and P is the vapor pressure of the pure solvent. The higher the mole fraction of the solute, the lower the mole fraction of the solvent. Therefore, the solution with the lowest vapor pressure is the one with the highest concentration of solute.
Colligative properties are those that depend on the number of particles dissolved in a solvent. The more particles in the solvent, the greater the effect the solute has on the properties of the solution. Examples of colligative properties include boiling point elevation, freezing point depression, vapor pressure depression, and osmotic pressure. What is the freezing point of 0.500 m I2 in cyclohexane (normal freezing point = 6.55°C)? Kf,cyclohexane = 20.2°C/m
−3.6°C
Colligative properties are those that depend on the number of particles dissolved in a solvent. The more particles in the solvent, the greater the effect the solute has on the properties of the solution. Examples of colligative properties include boiling point elevation, freezing point depression, vapor pressure depression, and osmotic pressure. Which aqueous solution has the lowest freezing point?
0.25 m AlCl3 The solution with the lowest freezing point is 0.25 m AlCl3. The equation for freezing point depression is T = i∙m∙Kf. Since all four solutions are aqueous, the value of Kf is the same. Therefore, the change in temperature (T) will be proportional to the product of the van't Hoff factor (i) and the molality (m). AlCl3: 4(0.25) = 1 NaCl: 2(0.35) = 0.70 glucose: 1(0.50) = 0.50 MgBr2: 3(0.30) = 0.90 AlCl3 has the greatest product, so it will have the largest T and the lowest freezing point.
Colligative properties are those that depend on the number of particles dissolved in a solvent. The more particles in the solvent, the greater the effect the solute has on the properties of the solution. Examples of colligative properties include boiling point elevation, freezing point depression, vapor pressure depression, and osmotic pressure. Which aqueous solution has the highest boiling point?
0.75 m NaCl The solution with the highest boiling point is 0.75 m NaCl. The equation for boiling point elevation is T = i∙m∙Kb. Since all four solutions are aqueous, the value of Kb is the same. Therefore, the change in temperature (T) will be proportional to the product of the van't Hoff factor (i) and the molality (m). NaCl: 2(0.75) = 1.5 AlCl3: 4(0.30) = 1.2 glucose: 1(1.0) = 1.0 MgBr2: 3(0.40) = 1.2 NaCl has the greatest product, so it will have the largest T and the highest boiling point.
Molality is a common unit of concentration which is used when temperature changes are a factor since its value will not change with temperature. Molality is calculated by determining the number of moles of the solute, or the substance being dissolved, in 1 kilogram of solvent. This is summarized as follows: molality equals moles of solute over 1 kilogram of solvent What is the molality of a solution prepared from 13.2 g MgCl2 and 175.0 g H2O?
0.794 m MgCl2 The molality of the solution is 0.794 m MgCl2. Calculate the moles of MgCl2 from the given mass. 13.2 grams MgCl2 times 1 mole MgCl2 over 95.21 grams MgCl2 equals 0.139 moles MgCl2 The molality would be calculated according to the following equation using the calculated moles of solute and the given mass of solvent, converted to kilograms: molality equals moles solute over kilograms of solvent equals 0.139 moles MgCl2 over 0.175 kilograms H2O which equals 0.794 molal MgCl2
Molarity is one of the most commonly used concentration units for solutions and is abbreviated as "M." Molarity is calculated by determining the number of moles of the solute, or the substance being dissolved, in 1 liter of solution. In other words, molarity is the number of moles/liter. This is summarized as follows: molarity equals moles of solute over 1 liter of solution What is the molarity of a solution prepared from 19.4 g NaCl in 300.0 mL of solution?
1.11 M NaCl The molarity of the solution is 1.11 M NaCl. Calculate the moles of NaCl from the given mass. 19.4 grams NaCl times 1 mole NaCl over 58.44 grams NaCl equals 0.332 moles NaCl The molarity would be calculated according to the following equation using the calculated moles and the given volume, in units of liters: molarity equals moles solute over liters of solution equals 0.332 moles of NaCl over 0.300 liters of solution, which equals 1.11 molar NaCl
Besides molarity (M), another frequently used concentration unit for solutions involves percent concentrations. Percent concentrations are calculated by determining the amount of solute, or the substance being dissolved, over the total solution, which is then multiplied by 100%. The total solution including both the solute and the solvent, which is the substance that does the dissolving, is summarized as follows: percent concentration equals amount of solute over amount of solution times 100 percent Another way to summarize the calculation is as follows: percent concentration equals amount of solute over the sum of amounts of solute and solvent times 100 percent There are three different types of percent concentrations, a) mass percent (m/m), b) volume percent (v/v), and c) mass/volume percent (m/v). The units for mass are normally in grams while the units for volume are normally in mL. What is the volume/volume percent of a solution prepared from 10.0 g ethanol in 95.0 g water? The density of ethanol is 0.789 g/mL and the density of water is 1.00 g/mL.
11.8% (v/v) ethanol The volume/volume percent of a solution containing 10.0 grams of ethanol (density = 0.789 g/mL) and 95.0 grams of water (density = 1.00 g/mL) is 11.8% (v/v) ethanol. Use the mass and density of each component to find their volumes. 10.0 grams ethanol times 1 milliliter ethanol over 0.789 grams equals 12.7 milliliters ethanol 95.0 grams H2O times 1 milliliter H2O over 1.00 gram equals 95.0 milliliters H2O Then, the volume/volume percent would be calculated according to the following equation: volume/volume percent equals 12.7 milliliters ethanol divided by the sum of 12.7 milliliters ethanol and 95.0 milliliters H2O times 100 percent volume/volume percent equals 11.8 percent by volume/volume ethanol
Besides molarity (M), another frequently used concentration unit for solutions involves percent concentrations. Percent concentrations are calculated by determining the amount of solute, or the substance being dissolved, over the total solution, which is then multiplied by 100%. The total solution including both the solute and the solvent, which is the substance that does the dissolving, is summarized as follows: percent concentration equals amount of solute over amount of solution times 100 percent Another way to summarize the calculation is as follows: percent concentration equals amount of solute over the sum of amounts of solute and solvent times 100 percent There are three different types of percent concentrations, a) mass percent (m/m), b) volume percent (v/v), and c) mass/volume percent (m/v). The units for mass are normally in grams while the units for volume are normally in mL. What is the volume/volume percent of a solution prepared from 19.5 mL methanol in 125 mL ethanol? Assume volumes are additive.
13.5% (v/v) methanol in ethanol The volume/volume percent of a solution containing 19.5 milliliters of methanol and 125 milliliters of ethanol is 13.5% (v/v) methanol in ethanol. The volume/volume percent would be calculated according to the following equation: volume/volume percent equals 19.5 milliliters methanol divided by the sum of 19.5 milliliters methanol and 125 milliliters ethanol times 100 percent volume/volume percent equals 13.5 percent by volume/volume methanol in ethanol
Besides molarity (M), another frequently used concentration unit for solutions involves percent concentrations. Percent concentrations are calculated by determining the amount of solute, or the substance being dissolved, over the total solution, which is then multiplied by 100%. The total solution including both the solute and the solvent, which is the substance that does the dissolving, is summarized as follows: percent concentration equals amount of solute over amount of solution times 100 percent Another way to summarize the calculation is as follows: percent concentration equals amount of solute over the sum of amounts of solute and solvent times 100 percent There are three different types of percent concentrations, a) mass percent (m/m), b) volume percent (v/v), and c) mass/volume percent (m/v). The units for mass are normally in grams while the units for volume are normally in mL. What is the mass/volume percent of a solution prepared from 35.0 g AgNO3 in 225 g solution? The density of the solution is 1.00 g/mL.
15.6% (m/v) AgNO3 The mass/volume percent of a solution containing 35.0 grams of AgNO3 and 225 grams of solution (density = 1.00 g/mL) is 15.6% (m/v) AgNO3. Use the mass and density of the solution to find its volume. 225 grams solution times 1 milliliter solution over 1.00 gram equals 225 milliliters solution Then, the mass/volume percent would be calculated according to the following equation: mass/volume percent equals 35.0 grams AgNO3 divided by 225 milliliter solution times 100 percent mass/volume percent equals 15.6 percent by mass/volume AgNO3 Note that you do not need to add the volume of the solute to the denominator because the amount of solution (solvent + solute) is given, not just the volume of solvent.
Besides molarity (M), another frequently used concentration unit for solutions involves percent concentrations. Percent concentrations are calculated by determining the amount of solute, or the substance being dissolved, over the total solution, which is then multiplied by 100%. The total solution including both the solute and the solvent, which is the substance that does the dissolving, is summarized as follows: percent concentration equals amount of solute over amount of solution times 100 percent Another way to summarize the calculation is as follows: percent concentration equals amount of solute over the sum of amounts of solute and solvent times 100 percent There are three different types of percent concentrations, a) mass percent (m/m), b) volume percent (v/v), and c) mass/volume percent (m/v). The units for mass are normally in grams while the units for volume are normally in mL. What is the mass percent of a solution prepared from 15.0 g NaCl and 75.0 g H2O?
16.7% (m/m) NaCl The mass percent of a solution containing 15.0 grams of NaCl in 75.0 grams of solution is 16.7% (m/m) NaCl. The mass percent would be calculated according to the following equation: mass percent equals 15.0 grams NaCl divided by the sum of 15.0 grams NaCl and 75.0 grams H2O times 100 percent mass percent equals 16.7 percent by mass NaCl
Besides molarity (M), another frequently used concentration unit for solutions involves percent concentrations. Percent concentrations are calculated by determining the amount of solute, or the substance being dissolved, over the total solution, which is then multiplied by 100%. The total solution including both the solute and the solvent, which is the substance that does the dissolving, is summarized as follows: percent concentration equals amount of solute over amount of solution times 100 percent Another way to summarize the calculation is as follows: percent concentration equals amount of solute over the sum of amounts of solute and solvent times 100 percent There are three different types of percent concentrations, a) mass percent (m/m), b) volume percent (v/v), and c) mass/volume percent (m/v). The units for mass are normally in grams while the units for volume are normally in mL. What is the volume percent (v/v) of a solution containing 25 mL of isopropyl alcohol, C3H7OH, dissolved in enough water to make a 150 mL solution?
17% (v/v) C3H7OH The volume percent of a solution containing 25 mL of isopropyl alcohol, C3H7OH, dissolved in enough water to make a 150 mL solution is 17% (v/v) C3H7OH. The volume percent would be calculated according to the following equation: volume percent (v/v) equals 25 milliliters C3H7OH over 150 milliliters of solution times 100 percent, so volume percent (v/v) equals 17 percent by volume C3H7OH
Molarity is one of the most commonly used concentration units for solutions and is abbreviated as "M." Molarity is calculated by determining the number of moles of the solute, or the substance being dissolved, in 1 liter of solution. In other words, molarity is the number of moles/liter. This is summarized as follows: molarity equals moles of solute over 1 liter of solution What is the molarity of a solution prepared from 85.0 g CaCl2 in 300.0 g of solution? The density of the solution is 1.05 g/mL.
2.68 M CaCl2 The molarity of a solution prepared from 85.0 grams CaCl2 in 300.0 grams of solution with a density of 1.05 g/mL is 2.68 M CaCl2. To find molarity, determine the moles of solute. 85.0 grams CaCl2 times 1 mole CaCl2 over 111.0 grams CaCl2 equals 0.766 moles CaCl2 Then, use density and mass to find the volume of the solution. 300.0 grams solution times 1 milliliter over 1.05 grams times 1 liter over 1000 milliliters equals 0.286 liters solution Finally, find the molarity using the calculated values. molarity equals 0.766 moles CaCl2 over 0.286 liters, so molarity equals 2.68 molar CaCl2
Colligative properties are those that depend on the number of particles dissolved in a solvent. The more particles in the solvent, the greater the effect the solute has on the properties of the solution. Examples of colligative properties include boiling point elevation, freezing point depression, vapor pressure depression, and osmotic pressure. What is the osmotic pressure of a 0.850 M solution of glucose in water at 35°C?
21.5 atm The osmotic pressure of a 0.850 M solution of glucose in water at 35°C is 21.5 atm. Use the formula for osmotic pressure with the ideal gas constant and the temperature in units of Kelvin to find, the osmotic pressure. Osmotic pressure equals the molar concentration times the ideal gas constant times temperature. Plug in values to find that the osmotic pressure equals 0.850 M times 0.0821 L atm per mol K times the quantity 35+273 Kelvin. Carry out the calculation to find that the osmotic pressure equals 21.5 atm.
Besides molarity (M), another frequently used concentration unit for solutions involves percent concentrations. Percent concentrations are calculated by determining the amount of solute, or the substance being dissolved, over the total solution, which is then multiplied by 100%. The total solution including both the solute and the solvent, which is the substance that does the dissolving, is summarized as follows: percent concentration equals amount of solute over amount of solution times 100 percent Another way to summarize the calculation is as follows: percent concentration equals amount of solute over the sum of amounts of solute and solvent times 100 percent There are three different types of percent concentrations, a) mass percent (m/m), b) volume percent (v/v), and c) mass/volume percent (m/v). The units for mass are normally in grams while the units for volume are normally in mL. What is the mass percent (m/m) of a solution containing 55 grams of calcium chloride, CaCl2, and 125 grams of water? Remember to select an answer with the correct number of significant figures.
31% (m/m) CaCl2 The mass percent of a solution containing 55 grams of calcium chloride, CaCl2, and 125 grams of water is 31% (m/m) CaCl2. The mass percent would be calculated according to the following equation: mass percent (m/m) equals 55 grams CaCl2 over sum of 125 grams H2O and 55 grams CaCl2 times 100 percent, so mass percent equals 31 percent by mass CaCl2
Molality is a common unit of concentration which is used when temperature changes are a factor since its value will not change with temperature. Molality is calculated by determining the number of moles of the solute, or the substance being dissolved, in 1 kilogram of solvent. This is summarized as follows: molality equals moles of solute over 1 kilogram of solvent What is the molality of a solution prepared from 25.0 g NaCl and 100.0 g H2O?
4.28 m NaCl The molality of the solution is 4.28 m NaCl. Calculate the moles of NaCl from the given mass. 25.0 grams NaCl times 1 mole NaCl over 58.44 grams NaCl equals 0.428 moles NaCl The molality would be calculated according to the following equation using the calculated moles of solute and the given mass of solvent, converted to kilograms: molality equals moles solute over kilograms of solvent equals 0.428 moles NaCl over 0.100 kilograms H2O which equals 4.28 molal NaCl
Molality is a common unit of concentration which is used when temperature changes are a factor since its value will not change with temperature. Molality is calculated by determining the number of moles of the solute, or the substance being dissolved, in 1 kilogram of solvent. This is summarized as follows: molality equals moles of solute over 1 kilogram of solvent What is the molality of a solution prepared from 80.3 g I2 and 100.0 mL hexane? The density of hexane is 0.655 g/mL.
4.82 m The molality of a solution prepared from 80.3 grams of I2 in 100.0 mL of hexane (density of hexane = 0.655 g/mL) is 4.82 m. To find molality, determine the moles of solute. 80.3 grams I2 times 1 mole I2 over 254.8 grams I2 equals 0.316 moles I2 Then, use density and mass to find the volume of the solution. 100.0 milliliters hexane times 0.655 grams hexane over 1 milliliter times 1 kilogram over 1000 grams equals 0.0655 kilograms hexane Finally, find the molality using the calculated values. molality equals 0.316 moles I2 over 0.0655 kilograms hexane, so molality equals 4.82 molal I2
Molarity is one of the most commonly used concentration units for solutions and is abbreviated as "M." Molarity is calculated by determining the number of moles of the solute, or the substance being dissolved, in 1 liter of solution. In other words, molarity is the number of moles/liter. This is summarized as follows: molarity equals moles of solute over 1 liter of solution What is the molarity of a solution containing 1.54 moles of NaOH and a volume of 2.5 × 102 mL?
6.2 M NaOH The molarity of a solution containing 1.54 moles of NaOH and a volume of 2.5 × 102 mL is 6.2 M NaOH. Since molarity is the number of moles in 1 liter, the volume must be converted from milliliters (mL) to liters (L). 250 milliliters over one times 1 liter over 1000 milliliters equals 0.25 liters The molarity would then be calculated according to the following equation: molarity equals 1.54 moles of NaOH over 0.25 liters which equals 6.2 moles over liter NaOH or 6.2 molar NaOH
Besides molarity (M), another frequently used concentration unit for solutions involves percent concentrations. Percent concentrations are calculated by determining the amount of solute, or the substance being dissolved, over the total solution, which is then multiplied by 100%. The total solution including both the solute and the solvent, which is the substance that does the dissolving, is summarized as follows: percent concentration equals amount of solute over amount of solution times 100 percent Another way to summarize the calculation is as follows: percent concentration equals amount of solute over the sum of amounts of solute and solvent times 100 percent There are three different types of percent concentrations, a) mass percent (m/m), b) volume percent (v/v), and c) mass/volume percent (m/v). The units for mass are normally in grams while the units for volume are normally in mL. What is the mass/volume percent of a solution prepared from 19.5 g NaOH in 125 g solution? The density of the solution is 1.05 g/mL.
6.4% (m/v) NaOH The mass/volume percent of a solution containing 19.5 grams of NaOH and 125 grams of solution (density = 1.05 g/mL) is 16.4% (m/v) NaOH. Use the mass and density of the solution to find its volume. 125 grams solution times 1 milliliter solution over 1.05 grams equals 119 milliliters solution Then, the mass/volume percent would be calculated according to the following equation: mass/volume percent equals 19.5 grams NaOH divided by 119 milliliter solution times 100 percent mass/volume percent equals 16.4 percent by mass/volume NaOH Note that you do not need to add the volume of the solute to the denominator because the amount of solution (solvent + solute) is given, not just the volume of solvent.
Colligative properties are those that depend on the number of particles dissolved in a solvent. The more particles in the solvent, the greater the effect the solute has on the properties of the solution. Examples of colligative properties include boiling point elevation, freezing point depression, vapor pressure depression, and osmotic pressure. What is the boiling point of 0.500 m I2 in cyclohexane (normal boiling point = 80.74°C)? Kb,cyclohexane = 2.79°C/m
82.14°C
Which substance is nonpolar?
C4H10 (butane) The nonpolar substance is butane (C4H10) because it has no net dipole moment. The remaining substances (H2O, CH3Cl, and C6H12O6) are all polar. Polarity results when a molecule has a net dipole moment resulting from the dipole moments of individual bonds not cancelling out. The dipole moment of the bond results from an uneven sharing of electrons.
A solution involves an attraction between the solute, the substance being dissolved, and the solvent, the substance that does the dissolving. This attraction is based on the polarity of the solute and the solvent. If both the solute and the solvent are polar, the solute will dissolve in the solvent. The same can also be said if the solute and the solvent are nonpolar. This results in the formation of a solution. However, if the solute and the solvent have differing polarities (i.e., one is polar and the other is nonpolar), they will not dissolve in one another, and a solution will not form. Which substance will dissolve in hexane?
CCl4 CCl4 will dissolve in hexane, as they are both nonpolar substances. All of the other substances (H2O, OF2, and CH2Cl2) are polar and will not be able to dissolve in hexane, due to differing polarities.
A solution involves an attraction between the solute, the substance being dissolved, and the solvent, the substance that does the dissolving. This attraction is based on the polarity of the solute and the solvent. If both the solute and the solvent are polar, the solute will dissolve in the solvent. The same can also be said if the solute and the solvent are nonpolar. This results in the formation of a solution. However, if the solute and the solvent have differing polarities (i.e., one is polar and the other is nonpolar), they will not dissolve in one another, and a solution will not form. Which substance will dissolve in hexane?
CCl4 CCl4 will dissolve in hexane, as they are both nonpolar substances. All of the other substances (H2O, OF2, and CH2Cl2) are polar and will not be able to dissolve in hexane, due to differing polarities.
A solution involves an attraction between the solute, the substance being dissolved, and the solvent, the substance that does the dissolving. This attraction is based on the polarity of the solute and the solvent. If both the solute and the solvent are polar, the solute will dissolve in the solvent. The same can also be said if the solute and the solvent are nonpolar. This results in the formation of a solution. However, if the solute and the solvent have differing polarities (i.e., one is polar and the other is nonpolar), they will not dissolve in one another, and a solution will not form. Which substance will dissolve in water?
CH3OH CH3OH will dissolve in water, as they are both polar substances. All of the other substances (CCl4, C4H10, BF3) are nonpolar and will not be able to dissolve in water, due to differing polarities. In addition, CH3OH can form hydrogen bonds with water and this ability increases its solubility in water.
Which substance is polar?
CH3OH (methanol) The polar substance is CH3OH because it has a net dipole moment. The remaining substances (Cl2, C6H6, and CO2) are all nonpolar. Polarity results when a molecule has a net dipole moment resulting from the dipole moments of individual bonds not canceling out. The dipole moment of the bond results from an uneven sharing of electrons. In addition, CH3OH can also form hydrogen bonds and this contributes to its ability to dissolve in water.
A solution involves an attraction between the solute, the substance being dissolved, and the solvent, the substance that does the dissolving. This attraction is based on the polarity of the solute and the solvent. If both the solute and the solvent are polar, the solute will dissolve in the solvent. The same can also be said if the solute and the solvent are nonpolar. This results in the formation of a solution. However, if the solute and the solvent have differing polarities (i.e., one is polar and the other is nonpolar), they will not dissolve in one another, and a solution will not form. Which substance will dissolve in ethanol?
H2O H2O will dissolve in ethanol, as they are both polar substances. All of the other substances (CCl4, benzene, and hexane) are nonpolar and will not be able to dissolve in ethanol, due to differing polarities.
Which substance is polar?
HCN The polar substance is HCN because it has a net dipole moment. The remaining substances (Br2, CS2, and C3H8) are all nonpolar. Polarity results when a molecule has a net dipole moment resulting from the dipole moments of individual bonds not cancelling out. The dipole moment of the bond results from an uneven sharing of electrons.
The ability for a substance to dissolve is affected by both temperature and pressure. Temperature As the temperature increases, the solubility of a solid solute also increases, as the solvent can hold more of the solute particles. However, as the temperature increases for a gas, the solubility decreases. As the temperature increases, the gas molecules have more energy and are able to escape the solvent, causing the solubility to decrease. Pressure As the pressure increases on a solution containing a gaseous solute, the solubility of the gas increases. At higher pressures, there are more gas molecules above the solution, which are able to enter and dissolve into the solution. Choose the true statement.
The solubility of a gas in water increases with an increase in pressure. The solubility of a gas in water increases with an increase in pressure and decreases with an increase in temperature.
The ability for a substance to dissolve is affected by both temperature and pressure. Temperature As the temperature increases, the solubility of a solid solute also increases, as the solvent can hold more of the solute particles. However, as the temperature increases for a gas, the solubility decreases. As the temperature increases, the gas molecules have more energy and are able to escape the solvent, causing the solubility to decrease. Pressure As the pressure increases on a solution containing a gaseous solute, the solubility of the gas increases. At higher pressures, there are more gas molecules above the solution, which are able to enter and dissolve into the solution. Choose the true statement.
The solubility of a gas in water increases with an increase in pressure. The solubility of a gas in water increases with an increase in pressure and decreases with an increase in temperature.
Osmosis involves a semi-permeable membrane with small pores that allow the solvent to travel across the membrane; however, larger solute molecules are unable to do so. The solvent is typically water. Water is able to diffuse across the membrane as it travels from the side with a lower solute concentration into the side with the higher solute concentration, in order to dilute the solute. The purpose of osmosis is to try and equalize the concentration of solute on both sides of the semi-permeable membrane. Assume there is a semipermeable membrane separating two compartments: the first compartment contains 3% glucose and the second compartment is 6% glucose. Select the answer that indicates what molecules will move and the correct direction.
Water (solvent) will move into the second compartment Water (solvent) will move into the second compartment to dilute the 6% glucose. Glucose is the solute and cannot travel across the membrane.