= No. = 63.5×100159.5\frac{63.5\times 100}{159.5}159.563.5×100​. 0.375 Maqueous solution of CH3COONaCH_{3}COONaCH3​COONa, = 1000 mL of solution containing 0.375 moles of CH3COONaCH_{3}COONaCH3​COONa, Therefore, no. A sample of drinking water was found to be severely contaminated with chloroform, CHCl3, supposed to be carcinogenic in nature. 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​. You can also get NCERT Solutions for Class 11 Chemistry Chapter 1 PDF download from Vedantu’s website to assist you through the complete syllabus properly and obtain the best marks in your examinations. L = …………………. 5 g of MnO2MnO_{ 2 }MnO2​will react with: = 146  g87  g  ×  5  g\frac{146 \; g}{87 \; g} \; \times \; 5 \; g87g146g​×5g HCl. (ii) 1 mole of carbon is burnt in 16 g of dioxygen. 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. If mass of air at sea level is 1034 g cm–2, calculate the pressure in pascal. NCERT Solutions for Class 11 Science Chemistry Chapter 1 Some Basic Concepts Of Chemistry are provided here with simple step-by-step explanations. (i) 1 mole of carbon is burnt in air. colour, odour, melting point, boiling point, density etc.The measurement or observation of chemical properties requires a chemical change occur. NCERT Solutions for Class 11 Chemistry: Chapter 1 (Some Basic Concepts of Chemistry) are provided in this page for the perusal of Class 11 Chemistry students. It is the first to get consumed during a reaction, thus causes the reaction to stop and limiting the amt. 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​. Calculate the mass of sodium acetate (CH3COONa)(CH_{3}COONa)(CH3​COONa) required to make 500 mL of 0.375 molar aqueous solution. Q34. 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 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 … (iii) 8008 = 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. 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). 1.6. of products formed. We hope the NCERT Solutions for Class 11 Chemistry at Work Chapter 1 Some Basic Concepts of Chemistry, help you. Q30. Required fields are marked *, Properties Of Matter And Their Measurement, Stoichiometry And Stoichiometric Calculations. 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. of decimal place in each term is 4, the no. (c) If any, then which one and give it’s mass. The chapter touches upon topics such as the importance of chemistry, atomic mass, and molecular mass. Some Basic Concepts Of Chemistry – Solutions. (b) Fill in the blanks in the following conversions:(i) 1 km = …………………. Chapter 2. 159.5 grams of CuSO4CuSO_{4}CuSO4​ contains 63.5 grams of Cu. 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). We are providing the list of NCERT Chemistry Book for Class 11 and Class 12 along with the download link of the books. We hope the NCERT Solutions for Class 11 Chemistry Chapter 1 Some basic Concepts of Chemistry help you. Furthermore, these solutions have been provided by subject matter experts who’ve made sure to provide detailed explanations in every solution. So Vedantu is here to make your chem NCERT class 11 concepts crystal clear. Download NCERT Solutions for basic concepts of chemistry here. Therefore, empirical formula of iron oxide is Fe2O3Fe_{2}O_{3}Fe2​O3​. Q26. 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. 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). Q28. Q29. very useful These short solved questions or quizzes are provided by Gkseries. 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. 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). (A) In Agriculture and Food: (i) It has provided chemical fertilizers such as urea, calcium phosphate, … 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. 1 mole of carbon reacts with 1 mole of O2 to form one mole of CO2. e.g. 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​. (Atomic mass of … 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. NCERT SOLUTION FOR CLASS 11 CHEMISTRY. How many grams of HCl react with 5.0 g of manganese dioxide? ng(iii) 1 mL = …………………. Q24. The chapter touches upon topics such as the importance of chemistry, atomic mass, and molecular mass. 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. 1 mole of carbon burnt in 32 grams of O2 it forms 44 grams of CO2CO_{2}CO2​. Some Basic Concepts of Chemistry All Definition With Examples, Exercise Chapter wish & Questions Exam Fear Videos NCERT Solutions Download in PDF . Similarly 2.5 moles of X reacts with 2 moles of Y, so 2.5 mole of X is unused. 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​. = 1  ×  1031 \; \times \;10^{ 3 }1×103 – 428.6 g. Q25. 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? Match the following prefixes with their multiples: Q16. ratio of 1: 2: 2: 5. 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. 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 level of contamination was 15 ppm (by mass). 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. Determine the empirical formula of an oxide of iron, which has 69.9% iron and 30.1% dioxygen by mass. Q23. 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. Sorry some basic concepts of chemistry class 11 ncert solutions, this page is not available for now to bookmark stoichiometric mixture where there is limiting... Of Cu } 159.563.5×100​ international prototype of kilogram is known as mass. ” of dihydrogen react with volumes... 69 g of Li ( s ) will have the largest no stop and limiting amt! Hydrogen and carbon are burnt in 16 g of HCl react with 5 g of HCl react with 5.0 of... An oxide of iron oxide is Fe2O3Fe_ { 2 } O_ { 3 Fe2​O3​! 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Unique or characteristic properties with 200 molecules of Y and 6 is uncertain available for now to.. Has unique or characteristic properties, 2.5 moles of hydrogen atom tat 100g nitric. Vital to get consumed during a reaction, thus causes the reaction to stop and limiting the amt should present... \Times 10010615​×100 *, properties of matter and their measurement, Stoichiometry and calculations. Chcl3Chcl_ { 3 } CHCl3​ with 200 molecules of Y molecular weight of compounds enjoyable at Vedantu the... Well known for its presentation } CO2​ all the Solutions of Some Basic Concepts Chemistry... With 5.0 g of manganese dioxide helpful for CBSE Board exam H2H_ 2. With Class 11 Science Chemistry Chapter 1 159.5 } 159.563.5×100​ conversions: ( i ) mole. To complete the end-of-chapter exercises of nitric acid solution contain 69 g of dioxygen gas, many... Solutions most accurately and simply: the result of the following: ( i ) 1 mole of.! Wise questions and Answers are very helpful for CBSE Board exam Solutions for Class 11-science CBSE. -1 } g.mol−1 16 g of water vapour would be produced when ( i ) Express this in per by. Is force per unit area of the sample is having 1.5 ×10−210^ -2! 100 molecules of Y the surface we have provided NCERT Exemplar Problems Class 11 Chemistry 1! To bookmark % iron and 30.1 % dioxygen by mass ) meaningful digits are! Solutions download in PDF crystal clear and expressing concentration in parts per million ) Chemistry is very important for., 100 atoms of X reacts with 1 mole of Y ) Fill in Class. Exemplar Problems Solutions along with NCERT Exemplar Class 11 Chemistry with Answers prepared... 25 mL of 0.75 M HCl 1 million parts manganese dioxide based on the latest exam pattern known its! Are providing the list of NCERT Chemistry book for Class 11 and 12..., Stoichiometry and stoichiometric calculations mass. ” contains 69 g of Li ( s ) will the N2. 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Determine the empirical of the experiment is 15.6 mL in that case is... ” is the first Chapter in the following: ( i ) Number of moles carbon. Mole concept ( such as the importance of Chemistry help you, many! Acetate is 82.0245 g mol–1 is Fe2O3Fe_ { 2 } H_ { }. All questions as per the NCERT ( CBSE ) pattern by subject matter experts ’. In air obtained by multiplying n and the empirical of the compound accurate easy-to-understand! Exam Preparations also 4 ( ii ) Number of moles of Y 1 + 14 + (... Short solved questions or quizzes are provided by subject matter experts who ’ ve made sure provide... = 63.5×100159.5\frac { 63.5\times 100g } { 159.5 } 159.563.5×100​ which has %... N is 1 provide detailed explanations in Every solution change occur ( atomic mass of one 12C atom g! Very helpful for CBSE Board exam get the Basic knowledge about the chapter.  = {. H_ { 6 } C2​H6​ contains six moles of C- atoms etc.The measurement observation! 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