Identify the limiting reagent, if any, in the following reaction mixtures. 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. These short solved questions or quizzes are provided by Gkseries. 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) 234,000 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. 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? The Class 11 Chemistry books of NCERT are very well known for its presentation. 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. Q19. 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. 1 atom of X reacts with 1 molecule of Y. In a reaction Similarly, 2.5 moles of X reacts with 2.5 moles of Y, so 2.5 mole of Y is unused. Q12. = 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 Q24. : The result of the experiment is 15.6 mL in that case 15 is certain and 6 is uncertain. (ii) 1 mole of carbon is burnt in 16 g of dioxygen. 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. Q8. 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. 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. 0.375 Maqueous solution of CH3COONaCH_{3}COONaCH3​COONa, = 1000 mL of solution containing 0.375 moles of CH3COONaCH_{3}COONaCH3​COONa, Therefore, no. Calculate the atomic mass (average) of chlorine using the following data: = [(Fractional abundance of 35Cl_{}^{35}\textrm{Cl}35​Cl)(molar mass of 35Cl_{}^{35}\textrm{Cl}35​Cl)+(fractional abundance of 37Cl_{}^{37}\textrm{Cl}37​Cl )(Molar mass of 37Cl_{}^{37}\textrm{Cl}37​Cl )], = [{(75.77100(34.9689u)\frac{75.77}{100}(34.9689u)10075.77​(34.9689u) } + {(24.23100(34.9659  u)\frac{24.23}{100}(34.9659\;u)10024.23​(34.9659u) }], Therefore, the average atomic mass of Cl = 35.4527 u, Q10. The chapter touches upon topics such as the importance of chemistry, atomic mass, and molecular mass. NCERT Solutions for Class 11 Chemistry … (iii) 8008 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 … easily explained 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. How many significant figures are present in the following? In three moles of ethane (C2H6), calculate the following: Classification of Elements and Periodicity in Properties. 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. 1 mol of MnO2MnO_{2}MnO2​ = 55 + 2 × 16 = 87 g, 1 mol of MnO2MnO_{2}MnO2​ reacts with 4 mol of HCl. Q3. Chapter 2. You can also get NCERT Solutions for Class 11 Chemistry Chapter 1 PDF download from … Therefore, 8.4 g of HCl will react with 5 g of MnO2MnO_{2}MnO2​. Q11. Significant figures indicate uncertainty in experimented value. 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 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. NCERT Solutions in Hindi medium have been created keeping those students in mind who are studying in a Hindi medium school. 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). Apart from these solutions, BYJU’S hosts some of the best subject experts who can guide the students to learn chemistry in a simplified and conceptual manner. The mass of O2 bear whole no. 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 … NCERT Solutions for Class 11 Science Chemistry Chapter 1 Some Basic Concepts Of Chemistry are provided here with simple step-by-step explanations. 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. Q26. 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. The remaining 18g of carbon (1.5 mol) will not undergo combustion. (c) 1 mole C2H6C_{2}H_{6}C2​H6​ contains six moles of H- atoms. Therefore, 16 grams of O2 will form 44×1632\frac{44\times 16}{32}3244×16​. Round up the following upto three significant figures: Q21. All the solutions of Some Basic Concepts of Chemistry - Chemistry explained in detail by experts to help students prepare for their CBSE exams. 1 mole of X reacts with 1 mole of Y. of significant numbers in the answer. Similarly, 200 atoms of X reacts with 200 molecules of Y, so 100 atoms of X are unused. 1 mole of carbon burnt in 32 grams of O2 it forms 44 grams of CO2CO_{2}CO2​. This is because the CBSE board question papers are set from the concepts covered in the NCERT books. . Express the following in the scientific notation: The chapter touches upon topics such as the importance of chemistry, atomic mass, and molecular mass. The molecular formula of a compound can be obtained by multiplying n and the empirical formula. They begin from Thomson’s model and move on to Rutherford’s and Bohr’s, successively disproving each one. 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 … of significant numbers in the answer is also 4. (iii) 2 moles of carbon are burnt in 16 g of dioxygen. 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 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%. Numerical problems in calculating mass percent and concentration. 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. 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. Weight of compounds for making 2.5 L of its 0.25 M solution the only elements of following... Mole concept ( such as the importance of Chemistry, atomic mass of the 11th NCERT Chemistry,. 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