The problem gives you all the information you need in order to solve for the molality and mole fraction of the solution. In order to determine its molarity, you're going to need the solution's volume.
To get the volume, you have to know what the density of the solution is. Determine the percent concentration by mass of the solution first
#"%w/w" = m_"solute"/m_"solution" * 100#
In your case, the mass of the solution will be
#m_"solution" = m_"glucose" + m_"water"#
#m_"solution" = 20 + 150 = "170 g"#
This means that you get
#"%w/w" = (20cancel("g"))/(170cancel("g")) * 100 = "11.8%"#
The density of this solution will thus be
//us.mt.com/us/en/home/supportive_content/application_editorials/D_Glucose_de_e.html
#rho = "1.045 g/mL"#
Use glucose's molar mass to determine how many moles you have
#20cancel("g") * "1 mole glucose"/(180.16cancel("g")) = "0.111 moles glucose"#
The solution's volume will be
#170cancel("g") * "1 mL"/(1.045cancel("g")) = "162.7 mL"#
This means that its molarity is - do not forget to convert the volume to liters!
#C = n/V = "0.111 moles"/(162.7 * 10^(-3)"L") = color(green)("0.68 M")#
A solution's molality is defined as the number of moles of solute divided by the mass of the solvent - in kilograms! This means that you have
#b = n/m_"water" = "0.111 moles"/(150 * 10^(-3)"kg") = color(green)("0.74 molal")#
To get the mole fraction of sucrose, you need to know how many moles of water you have present. Once again, use water's molar mass
#150cancel("g") * "1 mole water"/(18.02cancel("g")) = "8.24 moles water"#
The total number of moles the solution contains is
#n_"total" = n_"glucose" + n_"water"#
#n_"total" = 0.111 + 8.24 = "8.351 moles"#
This means that the mole fraction of sucrose, which is defined as the number of moles of sucrose divided by the total number of moles in the solution, will be
#chi_"sucrose" = n_"sucrose"/n_"total" = (0.111cancel("moles"))/(8.351cancel("moles")) = color(green)("0.013")#
SIDE NOTE I've left the values rounded to two sig figs, despite the fact that you only gave one sig fig for the mass of glucose.
Class-11-science»Chemistry
Some Basic Concepts of Chemistry
Dear Student,Moles of sucrose =Mass Molar mass =360342=1.05 moleMolarity =1.6 MMolarity =Number of moles of solute Volume in L1.6 M=1.05 moleVolume in LVolume in L =1.05 1.6 =0.657 L =0.657×1000 =657 mLDensity = Mass of ethanolVolume of ethanolMass of ethanol =0.89 ×657 =584.73 g
This page has all of the required homework for the material covered in the second exam of the second semester of General Chemistry. The textbook associated with this homework is CHEMISTRY The Central Science by Brown, LeMay, et.al. The last edition I required students to buy was the 12th edition (CHEMISTRY The Central Science, 12th ed. by Brown, LeMay, Bursten, Murphy and Woodward), but any edition of this text will do for this course.
Note: You are expected to go to the end of chapter problems in your textbook, find similar questions, and work out those problems as well. This is just the required list of problems for quiz purposes. You should also study the Exercises within the chapters. The exercises are worked out examples of the questions at the back of the chapter. The study guide also has worked out examples.
These are bare-bones questions. The textbook questions will have additional information that may be useful and that connects the problems to real life applications, many of them in biology.
- The density of ethanol is 0.789 g/mL and the density of water is 1.0 g/mL. A solution is made where 58.3 mL of ethanol, C2H5OH, is dissolved in 500 mL of water. Assume the volumes are additive.
- What is the mole fraction of ethanol in this solution? Answer
- What is the weight fraction? Answer
- What is the molarity? Answer
- What is the molality? Answer
( 58.3 mL ) ( ) ( ) = 1 mole C2H5OH ( 500 mL H2O ) ( ) ( ) = 27.8 mole H2O The total moles = 1 mole C2H5OH + 27.8 mole H2O = 28.8 moles
The mole fraction = = 1 mole C2H5OH 28.8 total moles = 0.035 The mole percent would be 3.5%.
( 58.3 mL ) ( ) = 46 g C2H5OH ( 500 mL H2O ) ( ) = 500 g H2O The total mass = 46 g C2H5OH + 500 g H2O = 546 g
The mass fraction = = 46 g C2H5OH 546 total grams = 0.084 The mass percent would be 8.4%.
The molarity = moles C2H5OH L of solution = = 1.79 M The molality = moles C2H5OH kg of solvent = = 2 m - Isopropyl alcohol at the drug store is usually 70% isopropyl alcohol, C3H7OH, and 30% water by volume. This corresponds to 54.94 g C3H7OH and 30 g H2O in 100 mL. The density of this solution is 0.8494 g/mL. In this problem assume water is the solute and the alcohol is the solvent.
- What is the mole fraction and mole percentage of water in this solution? Answer
- What is the mass fraction and mass percent of water? Answer
- What is the molarity of water? Answer
- What is the molality of water? Answer
( 54.94 g C3H7OH ) ( 1 mole C3H7OH 60 g C3H7OH ) = 0.916 mole C3H7OH ( 30 g H2O ) ( ) = 1.67 mole H2O The total moles = 0.916 mole C3H7OH + 1.67 mole H2O = 2.586 moles
The mole fraction = = 1.67 mole H2O 2.586 total moles = 0.646 The mole percent would be 64.6%.
The total mass = 54.94 g C3H7OH + 30 g H2O = 84.94 g
The mass fraction = = 30 g C2H2O 84.94 total grams = 0.353 The mass percent would be 35.3%.
To get the liters of solution we use the total mass and the density. The total mass is 54.94 g plus 30 g or 84.94 g.
( 84.94 g solution ) ( ) = 100 mL of solution ( 30 g H2O ) ( ) = 1.67 mole H2O The molarity = = = 16.7 M The molality = = 1.67 mole H2O 0.05494 kg C3H7OH = 30.4 m - What will the melting point and the boiling point be for a solution where 5 mL of hexane, C6H14, (density = 0.658 g/mL) is placed into 100 mL of benzene, C6H6 (density = 0.8765 g/mL)? The normal boiling point of benzene is 80.1 °C and the normal freezing point of benzene is 5.5 °C. For benzene the boiling point elevation constant, kb, is 2.53 °C/m and the freezing point depression constant, Kf, is 5.12 °C/m. Answer
( 5 mL C6H14 ) ( ) ( ) = 0.0383 mole C6H14 ( 100 mL C6H6 ) ( ) ( ) = 0.08765 kg C6H6
ΔTf = kfm =(5.12 °C/m)(0.437 m) = 2.24 °C New f.p. = 5.5 °C - 2.24 °C = 3.26 °C ΔTb = kbm =(2.53 °C/m)(0.437 m) = 1.1 °C New b.p. = 80.1 °C + 1.1 °C = 81.2 °CThe molality = 0.0383 mole C6H14 0.08765 kg C6H6 = 0.437 m - How many liters of ethylene glycol, C2H6O2, should be added to 3 L of water to make a solution that will freeze at -20 °C (-4 °F)? The density of C2H6O2 is 1.11 g/ml and the density of water is 1 g/ml. Answer ΔT = kfm ⇒ 20 °C = (1.86 °C/m)m ⇒
The molality = = 10.75 m = 10.75 moles C2H6O2 kg of water ( 3 L H2O ) ( ) ( 10.75 mole C2H6O2 1 kg H2O ) = 32.25 mole C2H6O2 ( 32.25 mole C2H6O2 ) ( 62 g C2H6O2 1 mole C2H6O2 ) ( 1 mL C2H6O2 1.11 g C2H6O2 ) = 1800 mL C2H6O2 = 1.8 L - List the following aqueous solutions in the order of increasing freezing point: 0.5 m Ca(NO3)2, 0.8 m sucrose, 0.6 m LiF. Answer
( 0.5 mole Ca(NO3)2 1 kg water ) ( 3 moles of particles 1 mole Ca(NO3)2 ) = 1.5 m in particles ( 0.8 mole sucrose 1 kg water ) ( 1 moles of particles 1 mole sucrose ) = 0.8 m in particles ( ) ( 2 moles of particles 1 mole LiF ) = 1.2 m in particles ΔTf = kfm suggests that the change in freezing point is directly related to the molality of the particles in the solution. The larger the molality, the larger the ΔT and the lower the freezing point. The highest freezing point will be the solution with the smallest molality, which is sucrose. The LiF solution will be next and the Ca(NO3)2 solution will have the lowest freezing point.
- Nandrolone is an anabolic steroid (a muscle-building chemical) which occurs naturally in the human body, but only in tiny quantities. A 20 g sample of nandrolone was placed in 500 mL of CCl4 (density = 1.59 g/mL) and the freezing point of the solution was found to be 1.75 °C lower than the normal freezing point. What is the molecular weight of nandrolone? kf for CCl4 is 29.8 °C/m. Answer
The molality = = = 0.0587 m = 0.0587 moles nandrolone kg of CCl4 (500 mL CCl4)(1.59 g/mL) = 795 g CCl4 = 0.795 kg CCl4
( 0.795 kg CCl4 ) ( 0.0587 mole nandrolone 1 kg CCl4 ) = 0.04667 mole nandrolone The molecular weight = 20 g nandrolone 0.04667 mole nandrolone = 428.5 g/mole (C28H44O3) - The partial pressure of oxygen, O2, at the highest recorded atmospheric pressure was 0.225 atm and the concentration in water under those conditions was 6.5 x 10-5 M. What was the concentration in water at the lowest recorded atmospheric pressure when the partial pressure of oxygen was 0.18 atm? Fish need 1.56 x 10-7 M O2 in order to survive. Do they have a high enough concentration at the lowest recorded atmospheric pressure? Answer Using Henry's Law: k = Cg/Pg = (6.5 x 10-5 M)/(0.225 atm) = 0.00029 M/atm. Under the new conditions the concentration would be: Cg = (0.00029 M/atm)(0.18 atm) = 5 x 10-5 M
The fish will have plenty of oxygen!
- The vapor pressure above a pure solvent is 120 torr. After a non-volatile solute is added to the solvent the pressure above the solution is 80 torr. What is the mole fraction of the solute in this solution? Answer
In this case the solute is non-volatile and so P°solute = 0.
Psoln = XsolventP°solvent + XsoluteP°solute = XsolventP°solvent
Xsolvent = Psoln / P°solvent = (80 torr)/(120 torr) = 0.667
Xsolvent + Xsolute = 1 ⇒ Xsolute = 1 - 0.667 = 0.333
-
Calculate the total vapor pressure above a solution at 20 °C when 100 moles of C6H12 are combined with 10 moles C7H16. Here is some data for these substances that can be used in this problem and the next two problems:
P° (at 20°C)DensityMolec.Wt. C6H12 77.7 Torr 0.779 g/mL 84.16 g/mole C7H16 40 Torr 0.684 g/mL 100.21 g/mole PC6H12 = ( 100 mole C6H12 110 Total Moles ) ( 77.7 Torr ) = 70.64 Torr (C6H12)
PTotal = 74.28 TorrPC7H16 = ( 10 mole C7H16 110 Total Moles ) ( 40 Torr ) = 3.636 Torr (C7H16) - What is the vapor pressure above a solution made by combining 500 mL C6H12 and 100 mL C7H16? See the data table in the previous problem. Answer
Get the moles of each:
( 500 mL C6H12 ) ( ) ( 1 mole C6H12 84.16 g C6H12 ) = 4.628 mole C6H12 ( 100 mL C7H16 ) ( ) ( 1 mole C6H12 100.21 g C7H16 ) = 0.6826 mole C7H16 PC6H12 = ( 4.628 mole C6H12 5.311 Total Moles ) ( 77.7 Torr ) = 67.7 Torr (C6H12)
PTotal = 72.84 TorrPC7H16 = ( 0.6826 mole C7H16 5.311 Total Moles ) ( 40 Torr ) = 5.14 Torr (C7H16) - What are the mole fractions of C6H12 and C7H16 if the total vapor pressure above the solution is 70 Torr? See the data table in the next to last problem. Answer Ptot = XC6H12P°C6H12 + XC7H16P°C7H16 ; 1 = XC6H12 + XC7H16 70 Torr = (1 - XC7H16)(77.7 Torr) + (XC7H16)(40 Torr) Solve to get XC7H16 = 0.2 and C6H12 = 0.8
- A mixture of sugar, C6H12O6, and 500 mL of water has a total vapor pressure of 9.0 torr at 10 °C. How much sugar was added to the water? At 10 °C the vapor pressure of pure water is 9.21 torr and the density of water is 0.9997 g/mL. Answer Ptot = XC6H12O6P°C6H12O6 + XH2OP°H2O ; P°C6H12O6 = 0 XH2O = Ptot/P°H2O = (9 torr)/(9.21 torr) = 0.977 = (mole H2O)/(total moles)
⇒ 0.977 = (27.77 mole H2O)/(x mole C6H12O6 + 27.77 mole H2O) Solve for x to get 0.654 mole C6H12O6 (0.654 mole C6H12O6)(180 g/mole) = 118 g C6H12O6( 500 mL H2O ) ( ) ( ) = 27.77 mole H2O - Describe the intermolecular forces for each of the solutions (a), (b), and (c) that would cause the behavior shown in each case.Answer
(a) An ideal solution, all intermolecular interactions are the same.
(b) Sovent-solvent and solute-solute intermolecular forces are greater than the solvent-solute intermolecular forces. It makes it harder to dissolve and more likely to go into the vapor phase.
(c) Solvent-solvent and solute-solue intermolecular forces are less than the solvent-solute intermolecular forces. The solute is attracted into the solvent and there is less vapor above the solution.