mass, Number of moles of solute (urea) = 11.11 g/ 60 g mol-1 solution = 0.5556 mol kg-1 and mole fraction of sugar Notes for Chemistry second Year of Federal Board, Multan Board, Faisalabad Board, Sargodha Board, Gujranwala Board, DG Khan Board, Rawalpindi Board or some other leading body of , Pakistan. cm-3. g x 6 + 1 g x 6 = 78 g mol-1, Molecular mass of carbon tetrachloride (CCl4) = Molality of solution = 2 molal = 2 mol mol kg-1, The mass of solvent (water) = 1 kg = 1000 g, Number of moles of solvent (water) = 1000/16 = 55.55, Mole fraction of solute = 2/(2 + 55.55) = 2/57.55 = 0.03475, Previous Topic: Numerical Problems on Molarity, Next Topic: Short Cut Methods to Calculate Concentration of Solution, Where you found this type of questions ? The level of problems given in this book is essentially required for JEE Main & Advanced aspirants. Calculate the molality of the solution. Given: Mass of solute = WB = 23 g, Molar mass of solute NCERT Exemplar Problems and Solutions for Class 12 Chemistry are available here for free download. nB) = 0.6795/4.8575 = 0.1399. Ans: The 12.68% and mole fraction of methyl alcohol is 0.0755 and that of water is Number of moles = ? solution is 7.690 mol kg-1 or 7.690 m. Calculate the mole fraction of solute in its 2 molal aqueous solution. moles x molecular mass = 4.22 x 98, Mass of solute (H2SO4) = 413.56 g = Complete Chemistry Notes For Fsc 2nd Year with questions,answers,mcqs,numerical problems,solutions,past paper,modal paper etc. NCERT Solutions for Class 6, 7, 8, 9, 10, 11 and 12. Find: Mole fraction of ethylene nB) = 0.1/10.1 = 0.0099, Ans: Molality of mol, Number of moles of NaOH = nB = 10 g/ 40 g = 0.25 10.0 g KCl is dissolved in 1000 g of water. 78 g = 0.3846 mol, Number of moles of solvent (carbon tetrachloride) = nB Molecular mass of ethylene glycol (C2H6O2) 0.5 + 3 = 3.5 mol, Mole fraction of solute (ethyl alcohol) = xB = nB/(nA + This book is an outcome of the experience gained during my interaction with the students going to appear in JEE Engineering Competition Exams. The chapter deals with the Group 15 elements, Oxides of Nitrogen, Allotropes of Phosphorus, Group 16 elements, simple oxides, allotropes of sulfur, chlorine, and other elements. Solution – For converting mass into mole and vice versa, we always need the molar mass. Your email address will not be published. 1.067 m, An aqueous solution of NaOH is marked 10% (w/w). This is the post on the topic of the 1st Year Chemistry Solved Exercise Numericals Chapter 1. Molecular mass of benzene (C6H6) = 12 CCl4), Mass of solvent (water) = 100 – 30 = 70 g, Number of moles of solute (benzene) = nB = 30 g/ The main motivation behind this page is to encourage understudies (particularly Pakistani understudies) to learn Chemistry. This chapter deals with crystalline and amorphous solids, as well as imperfections in solids. solution is 5.298 mol kg-1 or 5.298 m. The density of 5.35 M H2SO4 solution is 1.22 g cm-3. of solution in L = 0.2755/0.08347 = 3.301 M, Molality = Number of moles of solute/mass of sovent in kg, Molality = 0.2755 mol /0.073 kg = 3.774 mol L-1, Ans: The molarity of solution is 3.374 mol L-1 or 3.374 M, the molality of solution is 3.774 mol L-1 or 3.774 m. Calculate the mole fraction, molality and molarity of HNO3 in a solution containing 12.2 % HNO3. = 12 g x 2 + 1 g x 6 + 16g x 1 = 46 g mol-1, Molecular mass of acetic acid (CH3COOH) = 12 g x g/mL = 1.22 x 103 g/L = 1.22 kg/L, Mass of solution = Volume of solution x density = 1 L x Our chemistry problem solver online will help you to deal with any kind of assignment successfully. Calculate a) molality and b) molarity of the solution. Problem based on formula no. g + 25.2 g = 28.858 g, Percentage by mass = (Mass of solute/Mass of solution) x 100, ∴ Percentage by mass of urea = (3.658/28.858) x 100 = These old papers will assist understudies with preparing their paper as indicated by separate board. of solution = 5.35 M. Molecular mass H2SO4 = 1 g x 2 + Given: density of the solution = 1.22 g cm-3, Molarity Mass of glucose = 10 g and mass of H2O = 100 – 10 Nuclear decay 16. 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. = 0.25/5.25 = 0.0476, Volume of solution = Mass of solution / density = 100 g Mathway currently does not support this subject. It is comprised of 11 subtopics. Your email address will not be published. The chapter can be well understood through the exercises and solutions, that help to reinforce the fundamentals. 100 g = 0.1 kg, Molecular mass of urea (NH2CONH2) = 14 g x 2 + 1 g x 4 + 12 g x 1 + 16 g x 1, Molecular mass of urea (NH2CONH2) = and molality =? Ask our professionals to help you online and get instant feedback. Pls reply me, Past State Boards, CBSE, JEE and NEET Papers, Wooww, I love these questions They upgrade your thinking ability, Really its very helpful for us thank you so much, Your email address will not be published. 10.0 g KCl is dissolved in 1000 g of water. This is a collection of worked general chemistry and introductory chemistry problems, listed in alphabetical order.