Class 11: Chemistry Chapter 1 solutions. Complete Class 11 Chemistry Chapter 1 Notes.<\/p>\n\n\n\n
NCERT 11th Chemistry Chapter 1, class 11 Chemistry Chapter 1 solutions<\/p>\n\n\n\n
Page No: 22<\/p>\n\n\n\n
1.1. Calculate the molecular mass of the following :<\/strong><\/p>\n\n\n\n (i) H2<\/sub>O (ii) CO2 <\/sub>(iii) CH4<\/sub><\/strong><\/p>\n\n\n\n Answer<\/strong><\/p>\n\n\n\n (i) H2<\/sub>O = (2\u00d7Atomic mass of H) + (1\u00d7Atomic mass of O) (ii) CO2<\/sub> = (1 \u00d7 Atomic mass of C) + (2 \u00d7 Atomic mass of O) (iii) CH4 <\/sub>= (1 \u00d7 Atomic mass of carbon) + (4 \u00d7 Atomic mass of hydrogen) 1.2. Calculate the mass percent of different elements present in Sodium Sulphate (Na2<\/sub>SO4<\/sub>).<\/strong><\/p>\n\n\n\n Answer<\/strong><\/p>\n\n\n\n Molar mass of Na2<\/sub>SO4 <\/sub>= [(2 \u00d7 23.0) + (32.00) + 4 (16.00)] = 142 g<\/p>\n\n\n\n Mass percent of an element = (Mass of that element in compound\/Molar mass of that compound) \u00d7 100<\/p>\n\n\n\n \u2234 Mass percent of sodium (Na): (46\/142) \u00d7 100 = 32.39%<\/p>\n\n\n\n Mass percent of sulphur(S): (32\/142) \u00d7 100 = 22.54% NCERT 11th Chemistry Chapter 1, class 11 Chemistry Chapter 1 solutions<\/p>\n\n\n\n 1.3. Determine the empirical formula of an oxide of iron which has 69.9% iron and 30.1% dioxygen by mass.<\/strong><\/p>\n\n\n\n Answer<\/strong><\/p>\n\n\n\n % of iron by mass = 69.9 % [Given] Atomic mass of iron = 55.85 amu<\/p>\n\n\n\n Atomic mass of oxygen = 16.00 amu Simplest molar ratio = 1.25\/1.25 : 1.88\/1.25 <\/p>\n\n\n\n \u21d2 1 : 1.5 = 2 : 3<\/p>\n\n\n\n \u2234 The empirical formula of the iron oxide is Fe2<\/sub>O3<\/sub>.<\/p>\n\n\n\n 1.4. Calculate the amount of carbon dioxide that could be produced when Answer<\/strong><\/p>\n\n\n\n The balanced reaction of combustion of carbon in dioxygens is: (i) In dioxygen, combustion is complete. Therefore 1 mole of carbon dioxide produced by burning 1 mole of carbon. NCERT 11th Chemistry Chapter 1, class 11 Chemistry Chapter 1 solutions<\/p>\n\n\n\n 1.5. Calculate the mass of sodium acetate (CH3<\/sub>COONa) required to make 500 mL of 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245g mol-1<\/sup>.<\/strong><\/p>\n\n\n\n Answer<\/strong><\/p>\n\n\n\n 0.375 M aqueous solution of sodium acetate means that 1000 mL of solution containing 0.375 moles of sodium acetate.<\/p>\n\n\n\n \u2234No. of moles of sodium acetate in 500 mL = (0.375\/1000)\u00d7500 = 0.375\/2 = 0.1875<\/p>\n\n\n\n Molar mass of sodium acetate = 82.0245g mol-1<\/sup><\/p>\n\n\n\n \u2234Mass of sodium acetate acquired = 0.1875\u00d782.0245 g = 15.380g<\/p>\n\n\n\n 1.6. Calculate the concentration of nitric acid in moles per litre in a sample which has a density, 1.41 g mL-1<\/sup> and the mass per cent of nitric acid in it being 69%.<\/strong><\/p>\n\n\n\n Answer<\/strong><\/p>\n\n\n\n Mass percent of 69% means tat 100g of nitric acid solution contain 69 g of nitric acid by mass. 1.7. How much copper can be obtained from 100 g of copper sulphate (CuSO4<\/sub> )?<\/strong><\/p>\n\n\n\n Answer<\/strong><\/p>\n\n\n\n 1 mole of CuSO4<\/sub> contains 1 mole of copper. 1.8. 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.Given that the molar mass of the oxide is 159.69 g mol-1<\/sup><\/strong><\/p>\n\n\n\n Answer<\/strong><\/p>\n\n\n\n % of iron by mass = 69.9 % [Given] Atomic mass of iron = 55.85 amu<\/p>\n\n\n\n Atomic mass of oxygen = 16.00 amu Simplest molar ratio = 1.25\/1.25 : 1.88\/1.25 <\/p>\n\n\n\n \u21d2 1 : 1.5 = 2 : 3<\/p>\n\n\n\n \u2234 The empirical formula of the iron oxide is Fe2<\/sub>O3<\/sub>. Thus, Molecular formula is same as Empirical Formula i.e. Fe2<\/sub>O3<\/sub>.<\/p>\n\n\n\n NCERT 11th Chemistry Chapter 1, class 11 Chemistry Chapter 1 solutions<\/p>\n\n\n\n 1.9. Calculate the atomic mass (average) of chlorine using the following data :<\/strong><\/p>\n\n\n\n % Natural Abundance Molar Mass<\/strong><\/p>\n\n\n\n 35<\/sup>Cl 75.77 34.9689<\/strong><\/p>\n\n\n\n 37<\/sup>Cl 24.23 36.9659<\/strong><\/p>\n\n\n\n Answer<\/strong><\/p>\n\n\n\n Fractional Abundance of 35<\/sup>Cl = 0.7577 and Molar mass = 34.9689 1.10. In three moles of ethane (C2<\/sub>H6<\/sub>), calculate the following : Answer<\/strong><\/p>\n\n\n\n (i) 1 mole of C2<\/sub>H6 <\/sub> contains 2 moles of Carbon atoms 1.11. What is the concentration of sugar (C12<\/sub>H22<\/sub>O11<\/sub>) in mol L-1<\/sup> if its 20 g are dissolved in Answer<\/strong><\/p>\n\n\n\n Molar mass of sugar (C12<\/sub>H22<\/sub>O11<\/sub>) = (12\u00d712)+(1\u00d722)+(11\u00d716) = 342 g mol-1<\/sup> Volume of Solution = 2L (given)<\/p>\n\n\n\n Molar concentration = Moles of solute\/Volume of solution in L = 0.0585mol\/2L = 0.0293 mol L-1<\/sup> = 0.0293 M<\/p>\n\n\n\n 1.12. If the density of methanol is 0.793 kg L-1<\/sup> , what is its volume needed for making<\/strong> 2.5 L of its 0.25 M solution?<\/strong><\/p>\n\n\n\n Answer<\/strong><\/p>\n\n\n\n Molar mass of methanol (CH3<\/sub>OH) = (1\u00d712)+(4\u00d71)+(1\u00d716) = 32 g mol-1 <\/sup>= 0.032 kg mol-1<\/sup> Applying, M1<\/sub>V1<\/sub> (Given Solution) = M2<\/sub>V2 <\/sub>(Solution to be prepared) 1.13. Pressure is determined as force per unit area of the surface. The SI unit of pressure, pascal is as shown below : Answer<\/strong><\/p>\n\n\n\n Pressure is the force (i.e. weigh) acting per unit area. NCERT 11th Chemistry Chapter 1<\/p>\n\n\n\n 1.14. What is the SI unit of mass? How is it defined?<\/strong><\/p>\n\n\n\n Answer<\/strong><\/p>\n\n\n\n The SI unit of mass is kilogram (kg).<\/p>\n\n\n\n The kg is defined as the mass of platinum-iridium (Pt-Ir) cylinder that is stored in an air-tight jar at International Bureau of Weigh and Measures in France.<\/p>\n\n\n\n 1.15. Match the following prefixes with their multiples:<\/strong><\/p>\n\n\n\n Prefixes Multiples<\/strong><\/p>\n\n\n\n (i) micro 106<\/sup><\/strong><\/p>\n\n\n\n (ii) deca 109<\/sup><\/strong><\/p>\n\n\n\n (iii) mega 10-6<\/sup><\/strong><\/p>\n\n\n\n (iv) giga 10-15<\/sup><\/strong><\/p>\n\n\n\n (v) femto 10<\/strong><\/p>\n\n\n\n Answer<\/strong><\/p>\n\n\n\n Prefixes Multiples<\/p>\n\n\n\n (i) micro 10-6<\/sup><\/p>\n\n\n\n (ii) deca 10<\/p>\n\n\n\n (iii) mega 106<\/sup><\/p>\n\n\n\n (iv) giga 109<\/sup><\/p>\n\n\n\n (v) femto 10-15<\/sup><\/p>\n\n\n\n NCERT 11th Chemistry Chapter 1, class 11 Chemistry Chapter 1 solutions<\/p>\n\n\n\n 1.16. What do you mean by significant figures ?<\/strong><\/p>\n\n\n\n Answer<\/strong><\/p>\n\n\n\n Significant figures are meaningful digits which are known with certainty including the last digit whose value is uncertain.<\/p>\n\n\n\n For example, <\/p>\n\n\n\n In 11.2546 g, there are 6 significant figures but here 11.254 is certain and 6 is uncertain and the uncertainty would be \u00b11 in the last digit. Hence last uncertain digit is also included in Significant figures.<\/p>\n\n\n\n 1.17. A sample of drinking water was found to be severely contaminated with chloroform, CHCl3<\/sub>, supposed to be carcinogenic in nature. The level of contamination was 15 ppm (by mass). Answer<\/strong><\/p>\n\n\n\n (i) 15 ppm means 5 parts in million(106) <\/sup>parts. NCERT 11th Chemistry Chapter 1, class 11 Chemistry Chapter 1 solutions<\/p>\n\n\n\n 1.18. Express the following in the scientific notation: Answer<\/strong><\/p>\n\n\n\n (i) 0.0048 = 4.8\u00d710-3<\/sup> 1.19. How many significant figures are present in the following? Answer<\/strong><\/p>\n\n\n\n (i) 2 1.20. Round up the following upto three significant figures: Answer<\/strong><\/p>\n\n\n\n (i) 34.2 1.21. The following data are obtained when dinitrogen and dioxygen react together to form different compounds: (a) Fixing the mass of dinitrogen as 28 g, masses of dioxygen combined will be 32, 64, 32 and 80 g in the given four oxides. These masses of dioxygen bears a simple whole number ratio as 2:4:2:5. Hence, the data given will obey the law of multiple proportions. <\/p>\n\n\n\n The statement is as follows two elements always combine in a fixed mass of other bearing a simple ratio to another to form two or more chemical compounds.<\/p>\n\n\n\n (b) (i) 1 km = 1km\u00d71000m\/1km\u00d7100cm\/1m\/10mm\/1cm = 106 <\/sup>mm<\/p>\n\n\n\n 1 km = 1km\u00d71000m\/1km\u00d71pm\/10-12<\/sup>m = <\/sup>1015 <\/sup>pm<\/p>\n\n\n\n (ii) 1 mg = 1mg\u00d71g\/1000mg\u00d71kg\/1000g = 10-6<\/sup> kg<\/p>\n\n\n\n 1 mg = 1mg\u00d71g\/1000mg\u00d71ng\/10-9<\/sup>g = 10-6<\/sup> ng<\/p>\n\n\n\n (iii) 1 mL = 1mL\u00d71L\/1000mL = 10-3<\/sup> L<\/p>\n\n\n\n
= [2(1.0084) + 1(16.00)] amu = 2.016 u + 16.00 amu = 18.016 amu<\/p>\n\n\n\n
= [1(12.011) + 2 (16.00)] amu = 12.011 amu + 32.00 u = 44.01 amu<\/p>\n\n\n\n
= [1(12.011) + 4 (1.008)] amu = 12.011 amu + 4.032 amu = 16.043 amu<\/p>\n\n\n\n
Mass percent of oxygen:(O): (64\/142) \u00d7 100 = 45.07%<\/p>\n\n\n\n
% of oxygen by mass = 30.1 % [Given]<\/p>\n\n\n\n
Relative moles of iron in iron oxide = %mass of iron by mass\/Atomic mass of iron = 69.9\/55.85 = 1.25
Relative moles of oxygen in iron oxide = %mass of oxygen by mass\/Atomic mass of oxygen = 30.01\/16=1.88<\/p>\n\n\n\n
(i) 1 mole of carbon is burnt in air.
(ii) 1 mole of carbon is burnt in 16 g of dioxygen.
(iii) 2 moles of carbon are burnt in 16 g of dioxygen.<\/strong><\/p>\n\n\n\n
C(s) + O2<\/sub>(g) \u2192 CO2 <\/sub> (g)
1mole 1mole(32g) 1mole(44g)<\/p>\n\n\n\n
(ii) Here, oxgen acts as a limiting reagent as only 16 g of dioxygen is available. Hence, it will react with 0.5 mole of carbon to give 22 g of carbon dioxide.
(iii) Here again oxgen acts as a limiting reagent as only 16 g of dioxygen is available. It is a limiting reactant. Thus, 16 g of dioxygen can combine with only 0.5 mole of carbon to give 22 g of carbon dioxide.<\/p>\n\n\n\n
Molar mass of nitric acid (HNO3<\/sub>) = 1+14+48 = 63g mol-1<\/sup>
Number of moles in 69 g of HNO3 <\/sub>= 69\/63 moles = 1.095 moles
Volume of 100g nitric acid solution = 100\/1.41 mL = 70.92 mL = 0.07092 L
\u2234 Conc. of HNO3 <\/sub>in moles per litre = 1.095\/0.07092 = 15.44 M<\/p>\n\n\n\n
Molar mass of CuSO4 = (63.5) + (32.00) + 4(16.00)
= 63.5 + 32.00 + 64.00 = 159.5 g
159.5 g of CuSO4 contains 63.5 g of copper.
\u2234 copper can be obtained from 100 g of copper sulphate = (63.5\/159.5)\u00d7100 = 39.81g<\/p>\n\n\n\n
% of oxygen by mass = 30.1 % [Given]<\/p>\n\n\n\n
Relative moles of iron in iron oxide = %mass of iron by mass\/Atomic mass of iron = 69.9\/55.85 = 1.25
Relative moles of oxygen in iron oxide = %mass of oxygen by mass\/Atomic mass of oxygen = 30.01\/16=1.88<\/p>\n\n\n\n
Mass of Fe2<\/sub>O3 = <\/sub>(2\u00d755.85) + (3\u00d716.00) = 159.7 g mol-1<\/sup>
n = Molar mass\/Empirical formula mass = 159.7\/159.6 = 1(approx)<\/p>\n\n\n\n
Fractional Abundance of 37<\/sup>Cl = 0.2423 and Molar mass = 36.9659
\u2234 Average Atomic mass = (0.7577\u00d734.9689)amu + (0.2423\u00d736.9659)
= 26.4959 + 8.9568 = 35.4527<\/p>\n\n\n\n
(i) Number of moles of carbon atoms.
(ii) Number of moles of hydrogen atoms.
(iii) Number of molecules of ethane.<\/strong><\/p>\n\n\n\n
\u2234 3 moles of of C2<\/sub>H6 <\/sub>will contain 6 moles of Carbon atoms
(ii) 1 mole of C2<\/sub>H6 <\/sub> contains 6 moles of Hydrogen atoms
\u2234 3 moles of of C2<\/sub>H6 <\/sub>will contain 18 moles of Hydrogen atoms
(iii) 1 mole of C2<\/sub>H6 <\/sub> contains Avogadro’s no. 6.02\u00d71023 <\/sup>molecules
\u2234 3 moles of of C2<\/sub>H6 <\/sub>will contain ethane molecule = 3\u00d76.02\u00d71023<\/sup>= 18.06\u00d71023 <\/sup>molecules.<\/p>\n\n\n\n
enough water to make a final volume up to 2L?<\/strong><\/p>\n\n\n\n
No. of moles in 20g of sugar = 20\/342 = 0.0585 mole<\/p>\n\n\n\n
Molarity of the solution = 0.793\/0.032 = 24.78 mol L-1 <\/sup><\/p>\n\n\n\n
24.78\u00d7V1 <\/sub>= 0.25\u00d72.5 L
V1<\/sub>= 0.02522 L = 25.22 mL<\/p>\n\n\n\n
1Pa = 1N m-2<\/sup>
If mass of air at sea level is 1034 g cm-2<\/sup>,calculate the pressure in pascal.<\/strong><\/p>\n\n\n\n
P= F\/A = 1034g\u00d79.8ms-2<\/sup>\/cm2<\/sup>
=1034g\u00d79.8ms-2<\/sup>\/cm2 <\/sup>\u00d7 1kg\/1000g \u00d7 100cm\/1m \u00d7 100cm\/1m = 1.01332\u00d7105<\/sup> N
Now,
1Pa = 1N m-2<\/sup>
\u2234 1.01332\u00d7105<\/sup> N\u00d7m-2<\/sup> =1.01332\u00d7105<\/sup> Pa<\/p>\n\n\n\n
(i) Express this in percent by mass.
(ii) Determine the molality of chloroform in the water sample.<\/strong><\/p>\n\n\n\n
\u2234 % by mass = 15\/106 <\/sup>\u00d7 100 = 15\u00d710-4<\/sup> = 1.5\u00d710-3 <\/sup>%
(ii) Molar mass of chloroform(CHCl3<\/sub>) = 12+1+(3\u00d735.5) = 118.5 g mol-1<\/sup>
100g of the sample contain chloroform = 1.5\u00d710-3<\/sup>g
\u2234 1000 g(1 kg) of the sample will contain chloroform = 1.5\u00d710-2 <\/sup>g
= 1.5\u00d710-2<\/sup>\/118.65 mole = 1.266\u00d710-4 <\/sup>mole
\u2234 Molality = 1.266\u00d710-4 <\/sup>m.<\/p>\n\n\n\n
(i) 0.0048
(ii) 234,000
(iii) 8008
(iv) 500.0
(v) 6.0012<\/strong><\/p>\n\n\n\n
(ii) 234, 000 = 2.34\u00d7105<\/sup>
(iii) 8008 = 8.008\u00d7103<\/sup>
(iv) 500.0 = 5.000\u00d7102<\/sup>
(v) 6.0012 = 6.0012\u00d7100<\/sup><\/p>\n\n\n\n
(i) 0.0025
(ii) 208
(iii) 5005
(iv) 126,000
(v) 500.0
(vi) 2.0034<\/strong><\/p>\n\n\n\n
(ii) 3
(iii) 4
(iv) 3
(v) 4
(vi) 5<\/p>\n\n\n\n
(i) 34.216
(ii) 10.4107
(iii) 0.04597
(iv) 2808<\/strong><\/p>\n\n\n\n
(ii) 10.4
(iii) 0.046
(iv) 2810<\/p>\n\n\n\n
Mass of dinitrogen Mass of dioxygen
(i) 14 g 16 g
(ii) 14 g 32 g
(iii) 28 g 32 g
(iv) 28 g 80 g
(a) Which law of chemical combination is obeyed by the above experimental data?Give its statement.
(b) Fill in the blanks in the following conversions:
(i) 1 km = …………………. mm = …………………. pm
(ii) 1 mg = …………………. kg = …………………. ng
(iii) 1 mL = …………………. L = …………………. dm3<\/sup><\/strong><\/p>\n\n\n\n
Answer<\/strong><\/p>\n\n\n\n