= 90 g = 0.090 kg, Molecular mass NaOH = 23 g x 1 + 16 g x 1 + 1 g  x 1 = Ionic/Covalent Bonds 12. solution, Mass of solvent (water) = 100 – 20 = 80 g, Number of moles of solute (ethylene glycol) = nB Designed in a fairly simple language, we have made sure that the solutions cover all the fundamental concepts. mole fraction of ethyl alcohol and water in solution. nC = 1.3889 + 0.5435 + 0.8333 = 2.7657, Mole fraction of water = xA = nA/(nA +nB + nC) = 0.5435/2.7657 = 0.1965, Mole fraction of acetic acid = xC = nC/(nA +nB + nC) = 0.8333/2.7657 = 0.3013, Previous Topic: Numerical Problems on Perventage by Mass and Volume, Next Topic: Numerical Problems on Molarity, Your email address will not be published. The chapter deals with the properties of transition elements, colored ions and complex compounds, variation in sizes of elements, ionization enthalpies, magnetic properties, and oxidation states. Given: mass of solute (KCl) = 7.45 g, mass of solvent (water) = Given atomic masses H = 1, N = 14 and O = 16. Students learn the concepts of IUPAC nomenclature of Haloalkanes and Haloarenes. 18 g mol-1 = 3 mol, Total number of moles = nA + nB = 2 Chemistry: Matter and Change Solving Problems: A Chemistry Handbook SOLVING PROBLEMS: CHAPTER 1 A CHEMISTRY HANDBOOK Matter is made up of particles, called atoms, that are so small they cannot be seen with an ordinary light microscope. 5.189 mol, Number of moles of HNO3 = nB = 6.3 g/ This is the post on the topic of the 1st Year Chemistry Solved Exercise Numericals Chapter 1. = 0.1 mol, Molality = Number of moles of solute/Mass of solvent in kg. Mass of glucose = 10 g and mass of H2O = 100 – 10 0382, the molarity of solution is 2.011 mol L-1 or 2.011 M, the molality of solution is 2.206 mol kg-1 or 2.206 m. Calculate molarity and molality of 6.3 % solution of nitric acid having density 1.04 g cm-3. The objective behind this division is that students may revise the different types of problems being asked in Engineering Competition Entrance Exams such as JEE Main & Adavnced in an organised and progressive way. The students will learn the Kohlrausch Law and its applications. Get All The Latest Updates Delivered Straight Into Your Inbox For Free! mass of HNO3 = 6.3 %, Mass of HNO3 = 6.3 g and mass of H2O = 0.0556 mol, Volume of solution = Mass of solution / density = 100 g molecular mass, Number of moles of solute (methyl alcohol) = nB = 12.68% and mole fraction of methyl alcohol is 0.0755 and that of water is Given N = 14, H = 1, C = 12, O = 16. To Molar mass = 2+12+48 = 62g. Empirical and Molecular Formulas 7. H2O = 100 – 27 g = 73 g = 0.073 kg, Molecular mass H2SO4 = 1 g x 2 + They will also learn the applications of organometallic reactions, preparation of haloalkanes and haloarenes, and stereochemistry. 0.5 + 3 = 3.5 mol, Mole fraction of solute (ethyl alcohol) = xB = nB/(nA + Given: Mass of solute (sugar) = 34.2 g, Mass of solution (sugar Calculate a) molality and b) molarity of the solution. nB) = 1.2/1.5143 = 0.9245, Ans:  The percentage by mass of methyl alcohol is 100 – 6.3 g = 93.7 g = 0.0937 kg, Number of moles of water = nA = 93.4 g/ 18 g = 0.1143 + 1.4 = 1.5143 mol, Mole fraction of solute (methyl alcohol) = xB = nB/(nA + g = 90 g = 0.090 kg, Number of moles of glucose = nB = 10 g/ 180 g = /1.070 g cm-3 = 93.46 cm3 = 93.46 mL = 0.09346 L, Molarity of solution = Number of moles of the solute/volume Ask our professionals to help you online and get instant feedback. Each chapter in Numerical Problems in Physical Chemistry Book is divided into three levels of problems. In this article, we shall study numerical problems to calculate molality of a solution. Given Na = 23, H =1 , O = 16, Mass of NaOH = 10 g and mass of H2O = 100 – 10 g Formula:- Number of moles = Moles × N A. 4.444 + 0.3226 = 4.767 mol, Mole fraction of solute (ethylene glycol) = xB = Our chemistry problem solver online will help you to deal with any kind of assignment successfully. Calculate a) percentage by mass of methyl alcohol b) mole fraction of of solution in L = 0.0556/0.08333 =0.6672 M, Molality = 0.0556 mol /0.090 kg = 0.6178 mol kg-1. You may speak with a member of our customer support team by calling 1-800-876-1799. 100 g = 0.1 kg, Molecular mass of KCl = 39 g x 1 + 35.5 g x 1 = 74.5 g ” for JEE Main & Advanced Entrance Exams aspirants. /1.198 g cm-3 = 83.47 cm3 = 83.47 mL = 0.08347 L, Molarity of solution = Number of moles of the solute/volume NCERT textbooks are prescribed by CBSE as the best books for preparation of the school as well as board examinations. You may also browse chemistry problems according to the type of problem. 0.7952 g cm-3 = 3.658 g, Mass of solution = Mass of solute + Mass of solvent = 3.658 You will need to get assistance from your school if you are having problems entering the answers into your online assignment. Given: density of the solution = 1.04 g cm-3, % The chapter can be well understood through the exercises and solutions, that help to reinforce the fundamentals. Students also learn about the galvanic cell and the electrolytic cell. Calculate molarity, molality and mole fraction of NaOH in water. solution = 0.5556 mol kg-1 and mole fraction of sugar Solutions and Their Colligative Properties, Unsaturated solution Particles per million. moles x molecular mass = 5.35 x 98, Mass of solute (H2SO4) = 524.3 g = mass in aqueous solution. 16g  x 3 = 63 g mol-1, Number of moles of water = nA = 87.8 g/ 18 g = Molecular mass of ethylene glycol (C2H6O2) = 12 g x 12 + 1 g x 22 + 16 g x 11 = 342 g mol-1, Number of moles of solute (sugar) = nB = The NCERT Solutions provide ample material to enable students to form a good base with the fundamentals of the subject. molarity of solution is 1.852 mol L-1 and the molality What is the molality of H2SO4. = 0.8333 mol, Total number of moles = nA + nB + Given: 30% of benzene in carbon tetrachloride. of solution in L, Number of moles of solute = Molarity of solution They also deal with the preparation of alcohols, phenols, and ethers. 0.4545 + 0.3846 = 0.8391 mol, Mole fraction of solute (benzene) = xB = nB/(nA + This chapter also begins with the IUPAC nomenclature of alcohols, phenols and ethers. g/mL = 1.21 x 103 g/L = 1.21 kg/L, Mass of solution = Volume of solution x density = 1 L x 63 g = 0.1 mol, Volume of solution = Mass of solution / density = 100 g Catalysis 5. Given: density of the solution = 1.22 g cm-3, Molarity nB) = 0.1143/1.5143 = 0.0755, Mole fraction of solvent (water) = xA = nA/(nA + Important Numericals in Solution chapter | Physical Chemistry. This chapter is responsible from engineering and medical sciences point of view too. All the material distributed/posted on. Your email address will not be published. Old Papers for Chemistry FSc Part 2 of Federal Board, Multan Board, Faisalabad Board, Sargodha Board, Gujranwala Board, Rawalpindi Board or some other leading body of Punjab, Pakistan. This chapter deals with the kinetics, or the rate of a reaction. This chapter deals with the solid state. mass, Number of moles of solvent (water) = nB = 25.2 g/ nB) = 3/3.5 = 0.8571, Ans: Mole fraction of solute (ethyl alcohol) = 0.1429 and mole 40 g mol-1, Number of moles of water = nA = 90 g/ 18 g = 5 mass of sulphuric acid = 27%, Mass of H2SO4 = 27 g and mass of nC) = 1.3889/2.7657 = 0.5022, Mole fraction of ethyl alcohol = xB = nB/(nA +nB + Given H = 1, Cl = … 1 + 1 g x 4 + 16g  x 1 = 12 + 4 + 16 = 32 g mol-1, Number of moles of solute (methyl alcohol) = given mass/ Oxidation-Reduction Reactions 17. Mathematics XI (Punjab Text Book Board, Lahore)  These Mathematics-XI FSc Part 1 (first year) Notes are as indicated by "Punja... Class 12 Chemistry  Notes (Solutions), MCQs/Objective compose questions, demonstrate papers and old/past papers (of FBISE and BISE)... Chemistry  XII (Punjab Text Book Board, Lahore)  These Chemistry-XI FSc Part 2 (second year) Notes are as per "Punjab Text Book ... Computer Fundamentals (Sixth Edition)  Pradeep K. 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