Problems on empirical and molecular formulae. (c) 1 mole C2H6C_{2}H_{6}C2H6 contains six moles of H- atoms. Q26. Thus, the empirical of the given oxide is Fe2O3Fe_{2}O_{3}Fe2O3 and n is 1. = 63.5×100159.5\frac{63.5\times 100}{159.5}159.563.5×100. 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. 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. (ii) 1 mole of carbon is burnt in 16 g of dioxygen. 0.375 Maqueous solution of CH3COONaCH_{3}COONaCH3COONa, = 1000 mL of solution containing 0.375 moles of CH3COONaCH_{3}COONaCH3COONa, Therefore, no. Significant figures are the meaningful digits which are known with certainty. 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. Other problems related to the mole concept (such as percentage composition and expressing concentration in parts per million). The molecular formula of a compound can be obtained by multiplying n and the empirical formula. NCERT Solutions in Hindi medium have been created keeping those students in mind who are studying in a Hindi medium school. (iii) 8008 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. Similarly, 2.5 moles of X reacts with 2.5 moles of Y, so 2.5 mole of Y is unused. 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.”. Similarly, 200 atoms of X reacts with 200 molecules of Y, so 100 atoms of X are unused. Hence, 0.5 mol of Na2CO3Na_{ 2 }CO_{ 3 }Na2CO3 is in 1 L of water or 53 g of Na2CO3Na_{ 2 }CO_{ 3 }Na2CO3 is in 1 L of water. Q11. Match the following prefixes with their multiples: Q16. 1000 grams of the sample is having 1.5 ×10−210^{-2}10−2g of CHCl3CHCl_{3}CHCl3. Download NCERT Solutions for basic concepts of chemistry here. A + B2 → AB2 Calculate the mass per cent of different elements present in sodium sulphate (Na2SO4Na_{2}SO_{4}Na2SO4) . (b) Will the reactants N2 or H2 remain unreacted? 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. Q8. (i) 1 mole of carbon is burnt in air. Q27. = 1.5 ×10−2 gMolar mass of CHCl3\frac{1.5 \;\times 10^{-2} \;g}{Molar \; mass \; of \; CHCl_{3}}MolarmassofCHCl31.5×10−2g, Therefore, molality of CHCl3CHCl_{3}CHCl3 I water, Q18. 1 mole of carbon burnt in 32 grams of O2 it forms 44 grams of CO2CO_{2}CO2. 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. (iv) 500.0 (ii) Determine the molality of chloroform in the water sample. 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. 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. L = …………………. kg = …………………. 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. Calculate the mass of sodium acetate (CH3COONa)(CH_{3}COONa)(CH3COONa) required to make 500 mL of 0.375 molar aqueous solution. Now, No. = 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}Fe2O3, Empirical formula mass of Fe2O3Fe_{2}O_{3}Fe2O3, The molar mass of Fe2O3Fe_{2}O_{3}Fe2O3 = 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. Calculate the molar mass of the following: (i) CH4CH_{4}CH4 (ii)H2OH_{2}OH2O (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). Round up the following upto three significant figures: Q21. 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. mm = …………………. NCERT Solutions for Class 11 Chemistry … 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 … = 15106×100\frac{15}{10^{6}} \times 10010615×100. “Some Basic Concepts of Chemistry” is the first chapter in the Class 11 Chemistry syllabus as prescribed by NCERT. ≡ 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. NCERT Solutions for Class 11 Chemistry Some Basic Concepts of Chemistry Part 2 Class 11 Chemistry book solutions are available in PDF format for free download. Identify the limiting reagent, if any, in the following reaction mixtures. 1 mole of X reacts with 1 mole of Y. The subtopics covered under the chapter are listed below. We hope the NCERT Solutions for Class 11 Chemistry Chapter 1 Some basic Concepts of Chemistry help you. If the density of methanol is 0.793 kg L–1, what is its volume needed for making 2.5 L of its 0.25 M solution? NCERT Solutions for Class 11 Chemistry in PDF format for CBSE Board as well as UP Board updated for new academic year 2020-2021 onward are available to download free along with Offline Apps based on … (Atomic mass of … Since oxygen is the limiting reactant here, the 16g (0.5 mol) of O2 will react with 6g of carbon (0.5 mol) to form 22 g of carbon dioxide. The remaining 18g of carbon (1.5 mol) will not undergo combustion. Our expert professors of Chemistry explain the solutions of all questions as per the NCERT (CBSE) pattern. 1 mole of carbon reacts with 1 mole of O2 to form one mole of CO2. 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). Classification of Elements and Periodicity in Properties. of atoms. If mass of air at sea level is 1034 g cm–2, calculate the pressure in pascal. 1) Calculate the molecular mass of the following: i) H 2 O ii) CO 2 iii) CH 4 Solution The molecular mass of a compound is the sum of the atomic masses of the atoms present in the compound. Write bond-line formulas for: Isopropyl alcohol, 2,3-Dimethylbutanal, Heptan-4-one. of decimal place in each term is 4, the no. Answer Some Basic Concepts Of Chemistry – Solutions. 1.6. 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%. 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 … 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. Therefore, molecular formula is (CH)n(CH)_{ n }(CH)n that is C2H2C_{ 2 }H_{ 2 }C2H2. How many significant figures should be present in the answer of the following calculations? Q31. Q2. Furthermore, these solutions have been provided by subject matter experts who’ve made sure to provide detailed explanations in every solution. Class 11 Chemistry NCERT Solutions in English Medium: Class 11 Chemistry NCERT Solutions in Hindi Medium: Chapter 1 Some Basic Concepts of Chemistry: रसायन विज्ञान की कुछ मूल अवधारणाएँ: Chapter 2 Structure of The Atom: परमाणु की संरचना = 1197\frac{ 1 }{ 197 }1971 mol of Au (s), = 6.022 × 1023197\frac{ 6.022 \; \times \; 10^{ 23 } }{ 197 }1976.022×1023 atoms of Au (s), = 3.06 × 1021\times \; 10^{ 21 }×1021 atoms of Au (s), = 6.022 × 102323\frac{ 6.022 \; \times \; 10^{ 23 } }{ 23 }236.022×1023 atoms of Na (s), = 0.262 × 1023\times \; 10^{ 23 }×1023 atoms of Na (s), = 26.2 × 1021\times \; 10^{ 21 }×1021 atoms of Na (s), = 6.022 × 10237\frac{ 6.022 \; \times \; 10^{ 23 } }{ 7 }76.022×1023 atoms of Li (s), = 0.86 × 1023\times \; 10^{ 23 }×1023 atoms of Li (s), = 86.0 × 1021\times \; 10^{ 21 }×1021 atoms of Li (s), = 171\frac{ 1 }{ 71 }711 mol of Cl2Cl_{ 2 }Cl2 (g), (Molar mass of Cl2Cl_{ 2 }Cl2 molecule = 35.5 × 2 = 71 g mol−1mol^{ -1 }mol−1), = 6.022 × 102371\frac{ 6.022 \; \times \; 10^{ 23 } }{ 71 }716.022×1023 atoms of Cl2Cl_{ 2 }Cl2 (g), = 0.0848 × 1023\times \; 10^{ 23 }×1023 atoms of Cl2Cl_{ 2 }Cl2 (g), = 8.48 × 1021\times \; 10^{ 21 }×1021 atoms of Cl2Cl_{ 2 }Cl2 (g). (b) Fill in the blanks in the following conversions:(i) 1 km = …………………. We hope the NCERT Solutions for Class 11 Chemistry at Work Chapter 1 Some Basic Concepts of Chemistry, help you. Vedantu provides you with Class 11 Chemistry NCERT Solutions Chapter 1. Similarly, 100 atoms of X reacts with 100 molecules of Y. Some basic concepts of Chemistry Class 11 NCERT Solutions are given to make your study simplistic and enjoyable at Vedantu. (ii) Number of moles of hydrogen atom. Mole fraction of C2H5OHC_{ 2 }H_{ 5 }OHC2H5OH, = Number of moles of C2H5OHNumber of moles of solution\frac{Number \; of \; moles \; of \; C_{ 2 }H_{ 5 }OH}{Number \; of \; moles \; of \; solution}NumberofmolesofsolutionNumberofmolesofC2H5OH, 0.040 = nC2H5OHnC2H5OH + nH2O\frac{n_{C_{ 2 }H_{ 5 }OH}}{n_{C_{ 2 }H_{ 5 }OH} \; + \; n_{H_{ 2 }O}}nC2H5OH+nH2OnC2H5OH ——(1). 1Pa = 1N m–2 (c) Isopropyl alcohol. Class 11 Chemistry NCERT Solutions. 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. e.g. E.g. Molar mass of sodium acetate is 82.0245 g mol–1. 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. You can also get NCERT Solutions for Class 11 Chemistry Chapter 1 PDF download from Vedantuâs website to assist you through the complete syllabus properly and obtain the best marks in your examinations. Similarly, 2 moles of X reacts with 2 moles of Y, so 1 mole of Y is unused. 159.5 grams of CuSO4CuSO_{4}CuSO4 contains 63.5 grams of Cu. The mass of O2 bear whole no. 2 volumes of dihydrogen react with 1 volume of dioxygen to produce two volumes of vapour. = 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. (ii) 1 mole of carbon is burnt in 16 g of O2. (ii) 234,000 Pressure is determined as force per unit area of the surface. 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. Hence, Y is limiting agent. The chapter touches upon topics such as the importance of chemistry, atomic mass, and molecular 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. This makes the NCERT solutions provided by BYJU’S very student-friendly and concept-focused. NCERT Solutions for Class 11 Science Chemistry Chapter 1 - Some Basic Concepts Of Chemistry [FREE]. In a reaction Required fields are marked *, Properties Of Matter And Their Measurement, Stoichiometry And Stoichiometric Calculations. 1 mole of CuSO4CuSO_{4}CuSO4 contains 1 mole of Cu. = 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 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 Therefore, the given information obeys the law of multiple proportions. Hence, 10 volumes of dihydrogen will react with five volumes of dioxygen to produce 10 volumes of vapour. 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. Mass percent of 69% means tat 100g of nitric acid solution contain 69 g of nitric acid by mass. 5 g of MnO2MnO_{ 2 }MnO2will react with: = 146 g87 g × 5 g\frac{146 \; g}{87 \; g} \; \times \; 5 \; g87g146g×5g HCl. Significant figures indicate uncertainty in experimented value. Q20. ratio of 1: 2: 2: 5. (iii) 2 moles of carbon are burnt in 16 g of dioxygen. . You can also get NCERT Solutions for Class 11 Chemistry Chapter 1 PDF download from … It determines the extent of a reaction. Q3. Thus, 100 g of HNO3 contains 69 g of HNO3 by mass. NCERT SOLUTION FOR CLASS 11 CHEMISTRY. What do you mean by significant figures? (a) 1 mole C2H6C_{2}H_{6}C2H6 contains two moles of C- atoms. MCQ Questions for Class 11 Chemistry with Answers were prepared based on the latest exam pattern. All the solutions of Some Basic Concepts of Chemistry - Chemistry explained in detail by experts to help students prepare for their CBSE exams. These solutions can also be downloaded in a PDF format for free by clicking the download button provided below. (i) Number of moles of carbon atoms. How are 0.50 mol Na2CO3 and 0.50 M Na2CO3 different? These ncert book chapter wise questions and answers are very helpful for CBSE board exam. What is the SI unit of mass? 1 mole of X reacts with 1 mole of Y. (v) 6.0012. 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. A welding fuel gas contains carbon and hydrogen only. (a) 1 ppm = 1 part out of 1 million parts. Q19. (b) Heptan–4–one. Hence, X is limiting agent. Express the following in the scientific notation: 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). 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. 1 mol of MnO2MnO_{2}MnO2 = 55 + 2 × 16 = 87 g, 1 mol of MnO2MnO_{2}MnO2 reacts with 4 mol of HCl. 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. 1 atom of X reacts with 1 molecule of Y. 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 … These NCERT Solutions for Class 11 Chemistry Some Basic Concepts of Chemistry in Hindi medium pdf download have innumerable benefits as these are created in simple and easy-to-understand language.

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