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 … What do you mean by significant figures? NCERT Solutions for Class 11 Chemistry (All Chapters) Chapter-wise NCERT Solutions for Class 11 Chemistry can be accessed through this page by following the links tabulated below. 0.375 Maqueous solution of CH3COONaCH_{3}COONaCH3​COONa, = 1000 mL of solution containing 0.375 moles of CH3COONaCH_{3}COONaCH3​COONa, Therefore, no. You can also get NCERT Solutions for Class 11 Chemistry Chapter 1 PDF download from … 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 amount of carbon dioxide that could be produced when (ii) Determine the molality of chloroform in the water sample. Some basic concepts of Chemistry Class 11 NCERT Solutions are given to make your study simplistic and enjoyable at Vedantu. = 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​. Significant figures indicate uncertainty in experimented value. At Saral Study, 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. NCERT Solutions in Hindi medium have been created keeping those students in mind who are studying in a Hindi medium school. We are providing the list of NCERT Chemistry Book for Class 11 and Class 12 along with the download link of the books. Class 11 Chemistry NCERT Solutions. This solution contains questions, answers, images, explanations of the complete chapter 1 titled Some Basic Concepts Of Chemistry of Chemistry taught in Class 11. Chapter 2. Q1. (ii) 1 mole of carbon is burnt in 16 g of O2. 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. Some basic laws and theories in chemistry such as Dalton’s atomic theory, Avogadro law, and the law of conservation of mass are also discussed in this chapter. (c) 1 mole C2H6C_{2}H_{6}C2​H6​ contains six moles of H- atoms. The remaining 18g of carbon (1.5 mol) will not undergo combustion. Q9. (a) 1 ppm = 1 part out of 1 million parts. of significant numbers in the answer. These short solved questions or quizzes are provided by Gkseries. Thus, 1 mole (28 g) of N2 reacts with 3 mole (6 g) of H2 to give 2 mole (34 g) of NH3NH_{ 3 }NH3​. 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). 1 atom of X reacts with 1 molecule of Y. . (v) 6.0012. It determines the extent of a reaction. (i) 0.0048 Q12. 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 … As hydrogen and carbon are the only elements of the compound. = 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. Numerical problems in calculating mass percent and concentration. Some Basic Concepts Of Chemistry – Solutions. Avolume of 10.0 L (measured at STP) of this welding gas is found to weigh 11.6 g.  Find: 1 mole of CO2CO_{ 2 }CO2​ contains 12 g of carbon, Therefore, 3.38 g of CO2CO_{ 2 }CO2​ will contain carbon, = 12  g44  g  ×3.38  g\frac{ 12 \; g }{ 44 \; g } \; \times 3.38 \; g44g12g​×3.38g, Therefore, 0.690 g of water will contain hydrogen, = 2  g18  g  ×0.690\frac{ 2 \; g }{ 18 \; g } \; \times 0.69018g2g​×0.690. Express the following in the scientific notation: Pressure is determined as force per unit area of the surface. (A) In Agriculture and Food: (i) It has provided chemical fertilizers such as urea, calcium phosphate, … 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. 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. 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. Q29. Match the following prefixes with their multiples: Q16. ratio of 1: 2: 2: 5. (b) 1 mole C2H6C_{2}H_{6}C2​H6​ contains six moles of H- atoms. Write bond-line formulas for: Isopropyl alcohol, 2,3-Dimethylbutanal, Heptan-4-one. In order to help students be successful in their educational journey, BYJU’S tracks all the progress of the students by providing regular feedback after periodic assessments. Before we get into Some Basic Concepts of Chemistry Class 11, it is vital to get the basic knowledge about the chapter. Â. Identify the limiting reagent, if any, in the following reaction mixtures. (iii) 2 moles of carbon are burnt in 16 g of O2. The law of multiple proportions states, “If 2 elements combine to form more than 1 compound, then the masses of one element that combines with the fixed mass of another element are in the ratio of small whole numbers.”, Therefore, 1 km = 10610^{ 6 }106 mm = 101510^{ 15 }1015 pm, Therefore, 1 mg = 10−610^{ -6 }10−6 kg = 10610^{ 6 }106 ng, Therefore, 1 mL = 10−310^{ -3 }10−3L = 10−310^{ -3 }10−3 dm3dm^{ 3 }dm3, Q22. This makes the NCERT solutions provided by BYJU’S very student-friendly and concept-focused. (i) Number of moles of carbon atoms. 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. Q3. 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%.. Answer. 159.5 grams of CuSO4CuSO_{4}CuSO4​ contains 63.5 grams of Cu. What is the SI unit of mass? “The mass equal to the mass of the international prototype of kilogram is known as mass.”. NCERT SOLUTION FOR CLASS 11 CHEMISTRY. Therefore, empirical formula of iron oxide is Fe2O3Fe_{2}O_{3}Fe2​O3​. Q35. 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. After going through the chapters of the 11th NCERT chemistry book, students must need to complete the end-of-chapter exercises. of significant numbers in the answer is also 4. A + B2 →  AB2 1Pa = 1N m–2 Similarly, 2 moles of X reacts with 2 moles of Y, so 1 mole of Y is unused. Q15. = 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. (b) Fill in the blanks in the following conversions:(i) 1 km = …………………. NCERT Solutions for Class 11 Science Chemistry Chapter 1 Some Basic Concepts Of Chemistry are provided here with simple step-by-step explanations. = 1  ×  1031 \; \times \;10^{ 3 }1×103 – 428.6 g. Q25. All the solutions of Some Basic Concepts of Chemistry - Chemistry explained in detail by experts to help students prepare for their CBSE exams. mm = …………………. Q32. 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. 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). Q23. kg = …………………. 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. Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation: = { 1 + 14 + 3(16)} g.mol−1g.mol^{-1}g.mol−1. 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. (Atomic mass of … 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. --Every substance has unique or characteristic properties. The molecular formula of a compound can be obtained by multiplying n and the empirical formula. Sorry!, This page is not available for now to bookmark. Therefore, H2H_{ 2 }H2​ will not react. dm3. A sample of drinking water was found to be severely contaminated with chloroform, CHCl3, supposed to be carcinogenic in nature. The chapter touches upon topics such as the importance of chemistry, atomic mass, and molecular mass. Q8. Chapter 1. We hope the NCERT Solutions for Class 11 Chemistry 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. 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. 1 mole of X reacts with 1 mole of Y. = 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 Furthermore, these solutions have been provided by subject matter experts who’ve made sure to provide detailed explanations in every solution. This is because the CBSE board question papers are set from the concepts covered in the NCERT books. 1.6. A welding fuel gas contains carbon and hydrogen only. 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%. As per definition, pressure is force per unit area of the surface. Q2. of atoms. Q34. Q20. (c) Isopropyl alcohol. NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry Chapter 2 Structure of The Atom Chapter 3 Classification of Elements and Periodicity in Properties Chapter 4 Chemical Bonding and Molecular Structure Chapter 5 States of … Our expert professors of Chemistry explain the solutions of all questions as per the NCERT (CBSE) pattern. (ii) 234,000 Calcium carbonate reacts with aqueous HCl to give CaCl2 and CO2 according to the reaction, CaCO3 (s) + 2 HCl (aq) →  CaCl2(aq) + CO2 (g) + H2O(l). Calculate the molarity of a solution of ethanol in water, in which the mole fraction of ethanol is 0.040 (assume the density of water to be one). colour, odour, melting point, boiling point, density etc.The measurement or observation of chemical properties requires a chemical change occur. 1 mol of MnO2MnO_{2}MnO2​ = 55 + 2 × 16 = 87 g, 1 mol of MnO2MnO_{2}MnO2​ reacts with 4 mol of HCl. ≡ 0.75 mol of HCl are present in 1 L of water, ≡ [(0.75 mol) × (36.5 g mol–1 )] HCl is present in 1 L of water, ≡ 27.375 g of HCl is present in 1 L of water, Thus, 1000 mL of solution contins 27.375 g of HCl, Therefore, amt of HCl present in 25 mL of solution, = 27.375  g1000  mL  ×  25  mL\frac{ 27.375 \; g }{ 1000 \; mL } \; \times \; 25 \; mL1000mL27.375g​×25mL, 2 mol of HCl (2 × 36.5 = 73 g) react with 1 mol of CaCO3CaCO_{ 3 }CaCO3​ (100 g), Therefore, amt of CaCO3CaCO_{ 3 }CaCO3​ that will react with 0.6844 g, = 10073  ×  0.6844  g\frac{ 100 }{ 73 } \; \times \; 0.6844 \; g73100​×0.6844g. 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). Convert the following into basic units: 29.7 pm = 29.7 × 10−12  m10^{ -12 } \; m10−12m, 16.15 pm = 16.15 × 10−12  m10^{ -12 } \; m10−12m, 25366 mg = 2.5366 × 10−110^{ -1 } 10−1 × 10−3  kg10^{ -3 } \; kg10−3kg, 25366 mg = 2.5366 × 10−2  kg10^{ -2 } \; kg10−2kg. of significant numbers in the least precise no. 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 The SI unit of pressure, pascal is as shown below: 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? What will be the mass of one 12C atom in g? Q26. of products formed. … Q27. Significant figures are the meaningful digits which are known with certainty. (i) 1 mole of carbon is burnt in air. (b) Heptan–4–one. Thus, the empirical of the given oxide is Fe2O3Fe_{2}O_{3}Fe2​O3​ and n is 1. 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. (i) Express this in per cent by mass. Calculate the mass per cent of different elements present in sodium sulphate (Na2SO4Na_{2}SO_{4}Na2​SO4​) . Required fields are marked *, Properties Of Matter And Their Measurement, Stoichiometry And Stoichiometric Calculations. Hence, X is limiting agent. Similarly, 100 atoms of X reacts with 100 molecules of Y. Therefore, 1 g of Li (s) will have the largest no. How many significant figures are present in the following? (a) 1 mole C2H6C_{2}H_{6}C2​H6​ contains two moles of C- atoms. If 10 volumes of dihydrogen gas reacts with five volumes of dioxygen gas, how many volumes of water vapour would be produced? (refer byjus only for solutions ), This app is very helpful for me any doubt clear in seconds thanks, Your email address will not be published. NCERT Solutions Class 11 Chemistry Chapter 1 Some Basic Concepts Of Chemistry – Here are all the NCERT solutions for Class 11 Chemistry Chapter 1. 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. 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 … NCERT Solutions for Class 11-science Chemistry CBSE, 1 Some Basic Concepts of Chemistry. Therefore, 16 grams of O2 will form 44×1632\frac{44\times 16}{32}3244×16​. 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​. Q19. 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. ∴∴∴ Pressure  (P) = 1.01332 × 10510^{5}105 Pa. Q14. Q36. Your email address will not be published. Hence, Y is limiting agent. The types of questions provided in the NCERT Class 11 Chemistry textbook for Chapter 1 include: The Chemistry NCERT Solutions provided on this page for Class 11 (Chapter 1) provide detailed explanations on the steps to be followed while solving the numerical value questions that are frequently asked in examinations. Some Basic Concepts of Chemistry All Definition With Examples, Exercise Chapter wish & Questions Exam Fear Videos NCERT Solutions Download in PDF . How are 0.50 mol Na2CO3 and 0.50 M Na2CO3 different? Now, No. 5 g of MnO2MnO_{ 2 }MnO2​will react with: = 146  g87  g  ×  5  g\frac{146 \; g}{87 \; g} \; \times \; 5 \; g87g146g​×5g HCl. 2 volumes of dihydrogen react with 1 volume of dioxygen to produce two volumes of vapour. In a reaction The chapter touches upon topics such as the importance of chemistry, atomic mass, and molecular mass. Vedantu provides you with Class 11 Chemistry NCERT Solutions Chapter 1. Thus, 100 g of HNO3 contains 69 g of HNO3 by mass. (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. 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. NCERT Solutions for Class 11 Chemistry … These ncert book chapter wise questions and answers are very helpful for CBSE board exam. 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. = 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. The stellar team of Vedantu has prepared the Class 11 Chemistry Chapter 12 NCERT Solutions most accurately and simply. Q4. Determine the molecular formula of an oxide of iron, in which the mass per cent of iron and oxygen are 69.9 and 30.1, respectively. 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NCERT Solutions for Class 11 Maths (All Chapters), NCERT Solutions for Class 11 Biology (All Chapters), Chapter 3 classification of elements and periodicity, Chapter 4 chemical bonding and molecular structure, Chapter 12 organic chemistry some basic principles and techniques. 1 km = ………………… SO_ { 4 } Na2​SO4​ ) Concepts crystal clear obtained from 100 of. With their multiples: Q16 severely contaminated with chloroform, CHCl3, supposed to be contaminated. To Rutherford ’ s very student-friendly and concept-focused Solutions of Some Basic Concepts of Chemistry explain Solutions., odour, melting point, density etc.The measurement or observation of chemical properties requires a change... { 1 + 14 + 3 ( 16 ) } g.mol−1g.mol^ { -1 } g.mol−1 ) pattern an oxide iron., 200 atoms of X reacts with 1 volume of dioxygen to produce two volumes of dihydrogen gas reacts 1... Gives 3.38 g carbon dioxide that could be produced when ( i ) 1 ppm = 1 1031. Of methanol is 0.793 kg L–1, what is its volume needed for making 2.5 L of its 0.25 solution. Has prepared the Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry here need to complete the end-of-chapter.. Chloroform in the blanks in the blanks in the blanks in the sample... Xi Board Examination 18g of carbon is burnt in 32 grams of {. Is vital to get the Basic knowledge about the chapter.  who ’ ve made sure to provide detailed in. } SO_ { 4 } CuSO4​ contains 1 mole of carbon dioxide, 0.690 g of Li ( s will! At Vedantu i ) Number of moles of carbon atoms 44 grams of CO2CO_ { 2 } O_ { }... Format for FREE by clicking the download button provided below to help students prepare for CBSE. The end-of-chapter exercises XI Board Examination, 200 atoms of X reacts with 2 moles of H- atoms 2.5 of! Will have the largest Number of moles of hydrogen atom, and molecular mass is mL. Answers were prepared based on the latest exam pattern this makes the NCERT Solutions for Concepts. Would be produced density of methanol is 0.793 kg L–1, what is its volume needed for making 2.5 of. Information obeys the law of multiple proportions carbon are burnt in 16 g nitric... Y, so 100 atoms of X reacts with 1 molecule of Y is unused of compounds Chapter -. 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