Q1. 2 volumes of dihydrogen react with 1 volume of dioxygen to produce two volumes of vapour. Round up the following upto three significant figures: Q21. Q5. NCERT Solutions for Class 11th: Ch 1 Some Basic Concepts of Chemistry Ashutosh 03 Jun, 2015 NCERT Solutions for Class 11th: Ch 1 Some Basic Concepts of Chemistry NCERT Solutions for Class 11 Chemistry Chapter 1 – Some Basic Concepts of Chemistry In this chapter, laws of chemical combination, Dalton’s atomic theory, mole concept, empirical and molecular formula, stoichiometry and its calculations are discussed. After going through the chapters of the 11th NCERT chemistry book, students must need to complete the end-of-chapter exercises. (ii) Determine the molality of chloroform in the water sample. A welding fuel gas contains carbon and hydrogen only. NCERT Exemplar Class 11 Chemistry is very important resource for students preparing for XI Board Examination. Q9. A sample of drinking water was found to be severely contaminated with chloroform, CHCl3, supposed to be carcinogenic in nature. of significant numbers in the answer. Calculate the mass per cent of different elements present in sodium sulphate (Na2SO4Na_{2}SO_{4}Na2​SO4​) . Class 11 Chemistry NCERT Solutions. = 1.5  ×10−2  gMolar  mass  of  CHCl3\frac{1.5 \;\times 10^{-2} \;g}{Molar \; mass \; of \; CHCl_{3}}MolarmassofCHCl3​1.5×10−2g​, Therefore, molality of CHCl3CHCl_{3}CHCl3​ I water, Q18. NCERT Books chapter-wise Solutions (Text & Videos) are accurate, easy-to-understand and most helpful in Homework & Exam Preparations. = 63.5×100159.5\frac{63.5\times 100}{159.5}159.563.5×100​. Which one of the following will have the largest number of atoms? Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation: NCERT Solutions For Class 11 Chemistry Chapter 1: In CBSE Class 11 Chemistry Chapter 1, students will learn about the role played by chemistry in different dimensions of life.CBSE students who are looking for NCERT Solutions For Class 11 Chemistry … 1 mole of X reacts with 1 mole of Y. Hence, 0.5 mol of Na2CO3Na_{ 2 }CO_{ 3 }Na2​CO3​ is in 1 L of water or 53 g of Na2CO3Na_{ 2 }CO_{ 3 }Na2​CO3​ is in 1 L of water. ratio of 1: 2: 2: 5. Q35. (ii) 1 mole of carbon is burnt in 16 g of dioxygen. What mass of CaCO3 is required to react completely with 25 mL of 0.75 M HCl? (iii) 2 moles of carbon are burnt in 16 g of dioxygen. = 1034  g  ×  9.8  ms−2cm2×1  kg1000  g×(100)2  cm21m2\frac{1034\;g\;\times \;9.8\;ms^{-2}}{cm^{2}}\times \frac{1\;kg}{1000\;g}\times \frac{(100)^{2}\;cm^{2}}{1 m^{2}}cm21034g×9.8ms−2​×1000g1kg​×1m2(100)2cm2​, = 1.01332 × 10510^{5}105 kg m−1s−2m^{-1} s^{-2}m−1s−2, Pa   = 1 kgm−1kgm^{-1}kgm−1s−2s^{-2}s−2 Students can go through these Organic Chemistry Class 11 NCERT Solutions to learn the basics of organic chemistry along with some common terms used in this branch of chemistry. Hence, 10 volumes of dihydrogen will react with five volumes of dioxygen to produce 10 volumes of vapour. Q3. Some basic concepts of Chemistry Class 11 NCERT Solutions are given to make your study simplistic and enjoyable at Vedantu. Q20. pm(ii) 1 mg = …………………. Similarly, 100 atoms of X reacts with 100 molecules of Y. Therefore, empirical formula of iron oxide is Fe2O3Fe_{2}O_{3}Fe2​O3​. 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The molecular formula of a compound can be obtained by multiplying n and the empirical formula. Q2. Calculate the number of atoms in each of the following, 1 mole of Ar = 6.023  ×  10236.023 \; \times \; 10^{ 23 }6.023×1023 atoms of Ar, Therefore, 52 mol of Ar = 52 × 6.023  ×  10236.023 \; \times \; 10^{ 23 }6.023×1023 atoms of Ar, = 3.131  ×  10253.131 \; \times \; 10^{ 25 }3.131×1025 atoms of Ar, 1 u of He = 14\frac{ 1 }{ 4 }41​ atom of He, 52 u of He = 524\frac{ 52 }{ 4 }452​ atom of He, 4 g of He = 6.023  ×  10236.023 \; \times \; 10^{ 23 }6.023×1023 atoms of He, 52 g of He = 6.023  ×  1023  ×  524\frac{ 6.023 \; \times \; 10^{ 23 } \; \times \;52 }{ 4 }46.023×1023×52​ atoms of He, = 7.8286  ×  10247.8286 \; \times \; 10^{ 24 }7.8286×1024 atoms of He. Therefore, “the total number of digits in a number with the Last digit the shows the uncertainty of the result is known as significant figures.”. NCERT Solutions for Class 11-science Chemistry CBSE, 1 Some Basic Concepts of Chemistry. The NCERT solutions provided on this page contain step-by-step explanations in order to help students answer similar questions that might be asked in their examinations. = 1  ×  1031 \; \times \;10^{ 3 }1×103 – 428.6 g. Q25. If mass of air at sea level is 1034 g cm–2, calculate the pressure in pascal. NCERT Solutions Class 11 Chemistry Chapter 1 Some Basic Concepts Of Chemistry – Here are all the NCERT solutions for Class 11 Chemistry Chapter 1. ∴∴∴ Pressure  (P) = 1.01332 × 10510^{5}105 Pa. Q14. Q8. How is it defined? NCERT Chemistry Book for Class 11 and Class 12 are published by the officials of NCERT (National Council Of Educational Research and Training), New Delhi. NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry includes all the important topics with detailed explanation that aims to help students to understand the concepts better. We hope the NCERT Solutions for Class 11 Chemistry Chapter 1 Some basic Concepts of Chemistry help you. (ii) Number of moles of hydrogen atom. Q29. Therefore, 100 grams of CuSO4CuSO_{4}CuSO4​ will contain 63.5×100g159.5\frac{63.5\times 100g}{159.5}159.563.5×100g​ of Cu. They begin from Thomson’s model and move on to Rutherford’s and Bohr’s, successively disproving each one. easily explained Pressure is determined as force per unit area of the surface. ng(iii) 1 mL = …………………. Question from very important topics are covered by NCERT Exemplar Class 11.You also get idea about the type of questions and method to answer in your Class 11th … (i) 1 mole of carbon is burnt in air. Similarly, 2.5 moles of X reacts with 2.5 moles of Y, so 2.5 mole of Y is unused. dm3. Required fields are marked *, Properties Of Matter And Their Measurement, Stoichiometry And Stoichiometric Calculations. = 69.9055.85\frac{69.90}{55.85}55.8569.90​, = 1.251.25:1.881.25\frac{1.25}{1.25}:\frac{1.88}{1.25}1.251.25​:1.251.88​, Therefore, the empirical formula of oxide is Fe2O3Fe_{2}O_{3}Fe2​O3​, Empirical formula mass of Fe2O3Fe_{2}O_{3}Fe2​O3​, The molar mass of Fe2O3Fe_{2}O_{3}Fe2​O3​ = 159.69g, Therefore n = Molar  massEmpirical  formula  mass=159.69  g159.7  g\frac{Molar\;mass}{Empirical\;formula\;mass}=\frac{159.69\;g}{159.7\;g}EmpiricalformulamassMolarmass​=159.7g159.69g​. Hence, X is limiting agent. Therefore, the ratio of carbon to hydrogen is, Therefore, weight of 22.4 L of gas at STP, = 11.6  g10  L  ×  22.4  L\frac{ 11.6 \; g }{ 10 \; L } \; \times \; 22.4 \; L10L11.6g​×22.4L, n = Molar  mass  of  gasEmpirical  formula  mass  of  gas\frac{ Molar \; mass \; of \; gas}{Empirical \; formula \; mass \; of \; gas}EmpiricalformulamassofgasMolarmassofgas​, = 26  g13  g\frac{ 26 \; g }{ 13 \; g}13g26g​. In this NCERT solutions for class 11 chemistry class 11 NCERT solutions chapter, students learn all about atoms and the models introduced to represent their structure. The SI unit of pressure, pascal is as shown below: Mass percent of 69% means tat 100g of nitric acid solution contain 69 g of nitric acid by mass. Write bond-line formulas for : (a)2, 3–dimethyl butanal. 1Pa = 1N m–2 of atoms. Q34. Q31. Substituting the value of nH2On_{ H_{ 2 }O}nH2​O​ in eq (1), 0.96nC2H5OHn_{C_{ 2 }H_{ 5 }OH}nC2​H5​OH​ = 2.222 mol, = 2.314  mol1  L\frac{ 2.314 \; mol }{ 1 \; L }1L2.314mol​. Calculate the amount of carbon dioxide that could be produced when Here we are trying to give you a detailed answer to the questions of the entire topic of this chapter so that you can get more marks in your examinations by preparing the answers based on this … (A) In Agriculture and Food: (i) It has provided chemical fertilizers such as urea, calcium phosphate, … Your email address will not be published. The level of contamination was 15 ppm (by mass). The use of NCERT Books Class 11 Chemistry is not only suitable for studying the regular syllabus of various boards but it can also be useful for the candidates appearing for various competitive exams, Engineering Entrance Exams, and Olympiads. 1.6. Thus, the empirical of the given oxide is Fe2O3Fe_{2}O_{3}Fe2​O3​ and n is 1. NCERT Solutions for Class 11 Chemistry … (c) 1 mole C2H6C_{2}H_{6}C2​H6​ contains six moles of H- atoms. of significant numbers in the answer is also 4. (c) Isopropyl alcohol. Q19. At Saralstudy, we are providing you with the solution of Class 11th chemistry Some Basic Concepts of Chemistry according to the latest NCERT (CBSE) Book guidelines prepared by expert teachers. 1 mole of CuSO4CuSO_{4}CuSO4​ contains 1 mole of Cu. In three moles of ethane (C2H6), calculate the following: Hence, X is limiting agent. of products formed. (c) If any, then which one and give it’s mass. Hence, Y is limiting agent. --Every substance has unique or characteristic properties. The free Class 11 Chemistry NCERT solutions provided by BYJU’S have been prepared by seasoned academics and chemistry experts keeping the needs of Class 11 chemistry students in mind. = 15106×100\frac{15}{10^{6}} \times 10010615​×100. If the speed of light is 3.0 × 108 m s–1, calculate the distance covered by light in 2.00 ns, Speed of light = 3 × 10810^{ 8 }108 ms−1ms^{ -1 }ms−1, Distance travelled in 2 ns = speed of light * time taken, = (3 × 10810^{ 8 }108)(2 × 10−910^{ -9 }10−9). e.g. Here we are trying to give you a detailed answer to the questions of the entire topic of this chapter so that you can get more marks in your examinations by preparing the answers based on this lesson. Q30. (b) 1 mole C2H6C_{2}H_{6}C2​H6​ contains six moles of H- atoms. These Chemistry chapter 12 Class 11 NCERT Solutions are immensely beneficial from an exam point of view since they are based on CBSE pattern and will help you get high … 1 mole of X reacts with 1 mole of Y. Chlorine is prepared in the laboratory by treating manganese dioxide (MnO2) with aqueous hydrochloric acid according to the reaction: 4 HCl (aq) + MnO2(s) → 2H2O (l) + MnCl2(aq) + Cl2 (g). It is the first to get consumed during a reaction, thus causes the reaction to stop and limiting the amt. The chapter touches upon topics such as the importance of chemistry, atomic mass, and molecular mass. Similarly 2.5 moles of X reacts with 2 moles of Y, so 2.5 mole of X is unused. of moles of CH3COONaCH_{3}COONaCH3​COONa in 500 mL, = 0.3751000×500\frac{0.375}{1000}\times 50010000.375​×500, Molar mass of sodium acetate = 82.0245  g  mol−182.0245\;g\;mol^{-1}82.0245gmol−1, Therefore, mass that is required of CH3COONaCH_{3}COONaCH3​COONa, = (82.0245  g  mol−1)(0.1875  mole)(82.0245\;g\;mol^{-1})(0.1875\;mole)(82.0245gmol−1)(0.1875mole), Q6. Answer What will be the mass of one 12C atom in g? : The result of the experiment is 15.6 mL in that case 15 is certain and 6 is uncertain. = 0.793  kg  L−10.032  kg  mol−1\frac{0.793\;kg\;L^{-1}}{0.032\;kg\;mol^{-1} }0.032kgmol−10.793kgL−1​, Q13. of moles in 69 g of HNO3HNO_{3}HNO3​: = 69 g63 g mol−1\frac{69\:g}{63\:g\:mol^{-1}}63gmol−169g​, = Mass  of  solutiondensity  of  solution\frac{Mass\;of\;solution}{density\;of\;solution}densityofsolutionMassofsolution​, = 100g1.41g  mL−1\frac{100g}{1.41g\;mL^{-1}}1.41gmL−1100g​, = 70.92×10−3  L70.92\times 10^{-3}\;L70.92×10−3L, = 1.095 mole70.92×10−3L\frac{1.095\:mole}{70.92\times 10^{-3}L}70.92×10−3L1.095mole​, Therefore, Concentration of HNO3 = 15.44 mol/L. 159.5 grams of CuSO4CuSO_{4}CuSO4​ contains 63.5 grams of Cu. (iii) 8008 The NCERT solutions contain each and every exercise question asked in the NCERT textbook and, therefore, cover many important questions that could be asked in examinations. In a reaction Q15. E.g. Q32. (Atomic mass of … The subtopics covered under the chapter are listed below. Percent of Fe by mass = 69.9 % [As given above], Percent of O2 by mass = 30.1 % [As given above], = percent  of  iron  by  massAtomic  mass  of  iron\frac{percent\;of\;iron\;by\;mass}{Atomic\;mass\;of\;iron}Atomicmassofironpercentofironbymass​, = percent  of  oxygen  by  massAtomic  mass  of  oxygen\frac{percent\;of\;oxygen\;by\;mass}{Atomic\;mass\;of\;oxygen}Atomicmassofoxygenpercentofoxygenbymass​. NCERT Solution for Class 11 Chemistry Chapter 1: Some Basic Concepts of Chemistry In this chapter, students will be learn about the facts that the study of chemistry is significant as its domain encompasses every sphere of life. Amt of H2 = 1  ×  1031 \; \times \;10^{ 3 }1×103, 28 g of N2N_{ 2 }N2​ produces 34 g of NH3NH_{ 3 }NH3​, Therefore, mass of NH3NH_{ 3 }NH3​ produced by 2000 g of N2N_{ 2 }N2​, = 34  g28  g  ×  2000\frac{ 34 \; g }{ 28 \; g } \; \times \; 200028g34g​×2000 g. (b) H2H_{ 2 }H2​ is the excess reagent. = Number  of  moles  of  soluteVolume  of  solution  in  Litres\frac{Number\;of\;moles\;of\;solute}{Volume\;of\;solution\;in\;Litres}VolumeofsolutioninLitresNumberofmolesofsolute​, = Mass  of  sugarMolar  mass  of  sugar2  L\frac{\frac{Mass\;of\;sugar}{Molar\;mass\;of\;sugar}}{2\;L}2LMolarmassofsugarMassofsugar​​, = 20  g[(12  ×  12)  +  (1  ×  22)  +  (11  ×  16)]g]2  L\frac{\frac{20\;g}{[(12\;\times \;12)\;+\;(1\;\times \;22)\;+\;(11\;\times \;16)]g]}}{2\;L}2L[(12×12)+(1×22)+(11×16)]g]20g​​, = 20  g342  g2  L\frac{\frac{20\;g}{342\;g}}{2\;L}2L342g20g​​, = 0.0585  mol2  L\frac{0.0585\;mol}{2\;L}2L0.0585mol​, Therefore, Molar concentration = 0.02925 molL−1L^{-1}L−1. As hydrogen and carbon are the only elements of the compound. Molar mass of sodium acetate is 82.0245 g mol–1. (b) Fill in the blanks in the following conversions:(i) 1 km = …………………. If you have any query regarding NCERT Solutions for Class 11 Chemistry Chapter 1 Some basic Concepts of Chemistry, drop a comment below and we … If you have any query regarding NCERT Solutions for Class 11 Chemistry at Work Chapter 1 Some Basic Concepts of Chemistry, drop a comment below and we will get back to you at the earliest. Use the data given in the following table to calculate the molar mass of naturally occuring argon isotopes: = [(35.96755  ×  0.337100)( 35.96755 \; \times \; \frac{ 0.337 }{ 100 })(35.96755×1000.337​) + (37.96272  ×  0.063100)( 37.96272 \; \times \; \frac{ 0.063 }{ 100 })(37.96272×1000.063​) + (39.9624  ×  99.600100)( 39.9624 \; \times \; \frac{ 99.600 }{ 100 })(39.9624×10099.600​)], = [0.121 + 0.024 + 39.802] g  mol−1g \; mol^{ -1 }gmol−1, Q33. Calculate the molar mass of the following: (i) CH4CH_{4}CH4​      (ii)H2OH_{2}OH2​O      (iii)CO2CO_{2}CO2​, Molecular weight of methane, CH4CH_{4}CH4​, = (1 x Atomic weight of carbon) + (4 x Atomic weight of hydrogen), = (2 x Atomic weight of hydrogen) + (1 x Atomic weight of oxygen), Molecular weight of carbon dioxide, CO2CO_{2}CO2​, = (1 x Atomic weight of carbon) + (2 x Atomic weight of oxygen). Q17. Continue learning about various chemistry topics and chemistry projects with video lectures by visiting our website or by downloading our App from Google play store. Formula to calculate mass percent of an element = Mass  of  that  element  in  the  compoundMolar  mass  of  the  compound×100\frac{Mass\;of\;that\;element\;in\;the\;compound}{Molar\;mass\;of\;the\;compound}\times 100MolarmassofthecompoundMassofthatelementinthecompound​×100. The NCERT solutions that are provided here has been crafted with one sole purpose – to help students prepare for their examinations and clear them with good results. How are 0.50 mol Na2CO3 and 0.50 M Na2CO3 different? (b) Heptan–4–one. Therefore, H2H_{ 2 }H2​ will not react. = No. Burning a small sample of itin oxygen gives 3.38 g carbon dioxide, 0.690 g of water and no other products. (1) If we fix the mass of N2 at 28 g, the masses of N2 that will combine with the fixed mass of N2 are 32 grams, 64 grams, 32 grams and 80 grams. A + B2 →  AB2 We hope the NCERT Solutions for Class 11 Chemistry at Work Chapter 1 Some Basic Concepts of Chemistry, help you. Therefore, Mass percent of the sodium element: = 46.0g142.066g×100\frac{46.0g}{142.066g}\times 100142.066g46.0g​×100, = 32.066g142.066g×100\frac{32.066g}{142.066g}\times 100142.066g32.066g​×100, = 64.0g142.066g×100\frac{64.0g}{142.066g}\times 100142.066g64.0g​×100. (b) 100 grams of the sample is having 1.5 ×10−310^{-3}10−3g of CHCl3CHCl_{3}CHCl3​. Molar mass of Na2CO3Na_{ 2 }CO_{ 3 }Na2​CO3​: 1 mole of Na2CO3Na_{ 2 }CO_{ 3 }Na2​CO3​ means 106 g of Na2CO3Na_{ 2 }CO_{ 3 }Na2​CO3​, Therefore, 0.5 mol of Na2CO3Na_{ 2 }CO_{ 3 }Na2​CO3​, = 106  g1  mol  ×  0.5  mol\frac{ 106 \; g }{ 1 \; mol } \; \times \; 0.5 \; mol1mol106g​×0.5mol Na2CO3Na_{ 2 }CO_{ 3 }Na2​CO3​, 0.5 M of Na2CO3Na_{ 2 }CO_{ 3 }Na2​CO3​ = 0.5 mol/L Na2CO3Na_{ 2 }CO_{ 3 }Na2​CO3​. = 6.023  ×  10236.023 \; \times \; 10^{ 23 }6.023×1023 atoms of carbon, Therefore, mass of 1 12 C _{}^{ 12 }\textrm{ C }12​ C  atom, = 12  g6.022  ×  1023\frac{ 12 \; g }{ 6.022 \; \times \; 10^{ 23 }}6.022×102312g​, = 1.993  ×  10−23g1.993 \; \times \; 10^{ -23 } g1.993×10−23g. Therefore, 1 g of Li (s) will have the largest no. Students can note that the NCERT solutions provided by BYJU’S are free for all users to view online or to download as a PDF (which can be done by clicking the download button at the top of each chapter page). Calculate the concentration of nitric acid in moles per litre in a sample which has a density, 1.41 g mL–1 and the mass per cent of nitric acid in it being 69%. The following data are obtained when dinitrogen and dioxygen react together to form different compounds: (a) Which law of chemical combination is obeyed by the above experimental data?Give its statement. 1000 grams of the sample is having 1.5 ×10−210^{-2}10−2g of CHCl3CHCl_{3}CHCl3​. (i) Number of moles of carbon atoms. e.g. of significant numbers in the least precise no. These properties can be classified into twocategories – physical properties and chemical properties.Physical properties are those properties which can be measured or observed without changing the identity or the composition of the substance. These NCERT Solutions for Class 11 chemistry can help students develop a strong foundational base for all the important concepts included in the syllabi of both Class 11 as well as competitive exams. Q27. Furthermore, these solutions have been provided by subject matter experts who’ve made sure to provide detailed explanations in every solution. What is the concentration of sugar (C12H22O11) in mol L–1 if its 20 g are dissolved in enough water to make a final volume up to 2L? Determine the empirical formula of an oxide of iron, which has 69.9% iron and 30.1% dioxygen by mass. 1 mole of X reacts with 1 mole of Y. (a) 1 mole C2H6C_{2}H_{6}C2​H6​ contains two moles of C- atoms. (a) What is the mass of NH3NH_{ 3 }NH3​ produced if 2  ×  1032 \; \times \;10^{ 3 }2×103 g N2 reacts with 1  ×  1031 \; \times \;10^{ 3 }1×103 g of H2? Class 11 Chemistry Chapter 12 Organic Chemistry Some Basic Principles And Techniques Multiple Choice Questions and Answers for Board, JEE, NEET, AIIMS, JIPMER, IIT-JEE, AIEE and other competitive exams. Contains carbon and hydrogen only here we have provided NCERT Exemplar Problems Solutions along with download! 1: 2: 2: 2: 2: 2: 5 be carcinogenic in nature grams. With chloroform, CHCl3, supposed to be carcinogenic in nature nitric acid solution 69. Ncert books chapter-wise Solutions ( Text & Videos ) are accurate, and! 200 atoms of X reacts with 1 volume of dioxygen to produce 10 volumes of dihydrogen react 5. Level of contamination was 15 ppm ( by mass out of 1 some basic concepts of chemistry class 11 ncert solutions! This makes the NCERT ( CBSE ) pattern chloroform, CHCl3, supposed to be carcinogenic in nature {... Empirical of the sample is having 1.5 ×10−210^ { -2 } 10−2g of CHCl3CHCl_ 3. 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