Mole Concept Solved Examples

SOLVED EXAMPLES Ex. 1 Find the relative atomic mass, atomic mass of the following elements. (i) Na (i)i F (iii) H (iv) Ca (ii) 19, 19 amu (iii) 1, 1.008 amu , (iv) 40, 40 amu, (v) Ag Sol. (i) 23, 23 amu (v) 108, 108 amu. Ex. 2 14 g of Nitrogen gas and 22 g of CO2 gas are mixed together. Find the volume of gaseous mixture at STP. Sol. Moles of N2 = 14 = 0.5. 28 22 = 0.5. 44 So total moles = 0.5 + 0.5 = 1. So vol. at STP = 1 × 22.4 = 22.4 lit. moles of CO2 = Ex. 3 Sol. A sample of (C2H6) ethane has the same mass as 107 molecules of methane. How many C2H6 molecules does the sample contain ? 107 Moles of CH4 = NA Mass of CH4 = 107 × 16 = mass of C2H6 NA 10 7  16 NA  30 So Moles of C2H6 = So No. of molecules of C2H6 = 10 7  16 × NA = 5.34 × 106. NA  30 Ex. 4 From 160 g of SO2 (g) sample, 1.2046 x 1024 molecules of SO2 are removed then find out the volume of left over SO2 (g) at STP. Sol. Given moles = 160 = 2.5. 64 Removed moles = 1.2046  10 24 6.023  10 23 = 2. so left moles = 0.5. volume left at STP = 0.5 × 22.4 = 11.2 lit. Ex. 5 Show that in the reaction N2 (g) + 3H2(g)  2NH3 (g), mass is conserved. Sol. N2 (g) + 3H2(g)  2NH3 (g) moles before reaction 1 3 0 moles after reaction 0 0 2 Mass before reaction = mass of 1 mole N2(g) + mass of 3 mole H2(g) = 14 x 2 + 3 x 2 = 34 g mass after reaction = mass of 2 mole NH3 = 2 x 17 = 34 g. MOLE CONCEPT Ex. 6 Find the density of CO2(g) with respect to N2O(g). Sol. R.D. = Ex. 7 Find the vapour density of N2O5 Sol. V.D. = Ex. 8 Write a balance chemical equation for following reaction : M.wt.of CO 2 44 = = 1. M.wt.of N 2O 44 Mol. wt. of N2 O 5 = 54. 2 When ammonia (NH3) decompose into nitrogen (N2) gas & hydrogen (H2) gas. 1 3 N2 + H2 2 2 Sol. NH3  Ex. 9 When 170 g NH3 (M =17) decomposes how many grams of N2 & H2 is produced. Sol. NH3  or 2NH3  N2 + 3H2 . 1 3 N2 + H2 2 2 moles of H 2 moles of N 2 moles of NH 3 = = . 1/ 2 1 3/ 2 So moles of N2 = Similarly moles of H2 = So 1 170 × = 5. 2 17 So wt. of N2 = 5 × 28 = 140 g. 3 170 × = 15. 2 17 wt. of H2 = 15 × 2 = 30 g. Ex. 10 When x gram of a certain metal brunt in 1.5 g oxygen to give 3.0 g of its oxide. 1.20 g of the same metal heated in a steam gave 2.40 g of its oxide. shows the these result illustrate the law of constant or definite proportion Sol. Wt. of metal = 3.0 – 1.5 = 1.5 g so wt. of metal : wt of oxygen = 1.5 : 1.5 = 1 : 1 similarly in second case , wt. of oxygen = 2.4 – 1.2 = 1.2 g so wt. of metal : wt of oxygen = 1.2 : 1.2 = 1 : 1 so these results illustrate the law of constant proportion. Ex. 11 Find out % of O & H in H2O compound. Sol. % of O = 16 × 100 = 88.89% 18 % of H = 2 × 100 = 11.11% 18 Ex. 12 Ans. Acetylene & butene have empirical formula CH & CH2 respectively. The molecular mass of acetylene and butene are 26 & 56 respectively deduce their molecular formula. C2H2 & C4H8 Sol. n= Molecular mass Empirical formula mass For Acetylene : n= 26 =2 13  Molecular formula = C2H2 For Butene : 56 =4 14 Molecular formula = C4H8 . n=  Ex. 13 Ans. An oxide of nitrogen gave the following percentage composition : N = 25.94 and O = 74.06 Calculate the empirical formula of the compound. N2O5 Element Sol. % / Atomic mas s S imple ratio S imple intig er ratio N 25 . 94  1 .85 14 1 2 O 74.06  4.63 16 2.5 5 So empirical formula is N2O5. Ex. 14 340 g NH3 (M = 17) when decompose how many litres of nitrogen gas is produced at STP. Sol. NH3  1 3 N + H 2 2 2 2 moles of NH3 = So  Ex. 15 Sol. 340 = 20. 17 1 × 20 = 10. 2 vol. of N2 at STP = 10 × 22.4 = 224 lit. moles of N2 = 4 mole of MgCO3 is reacted with 6 moles of HCl solution. Find the volume of CO2 gas produced at STP, the reaction is MgCO3 + 2HCl  MgCl2 + CO2 + H2O. Here HCl is limiting reagent. So moles of CO2 formed = 3. So vol. at STP = 3 × 22.4 = 67.2 lit. Ex. 16 117 gm NaCl is dissolved in 500 ml aqueous solution. Find the molarity of the solution. Sol. Molarity = 117 / 58 .5 = 4M. 500 / 1000 MOLE CONCEPT Ex. 17 0.32 mole of LiAlH4 in ether solution was placed in a flask and 74 g (1 moles) of t-butyl alcohol was added. The product is LiAlHC12H27O3 . Find the weight of the product if lithium atoms are conserved. [Li = 7, Al = 27, H = 1, C = 12, O = 16] Sol. Applying POAC on Li 1 × moles of LiAlH4 = 1× moles of LiAlH C12H27O3 254 × 0.32 = 1 × wt. of LiAlH C12H27O3. wt. of LiAlH C12H27O3 = 81.28 gm. Ex. 18 Calculate the resultant molarity of following : (a) 200 ml 1M HCl + 300 ml water (b) 1500 ml 1M HCl + 18.25 g HCl (c) 200 ml 1M HCl + 100 ml 0.5 M H2SO4 (d) 200 ml 1M HCl + 100 ml 0.5 M HCl Ans. (a) 0.4 M (d) 0.83 M. Sol. (a) Final molarity = (b) 1.33 M (c) 1 M 200×1 + 0 = 0.4 M. 200 + 300 18.25 ×1000 36.5 = 1.33 M 1500 1500 ×1 + (b) Final molarity = (c) Final molarity of H+ = (d) Final molarity = Ex. 19 Ans. 200 ×1 +100 × 0.5× 2 = 1 M. 200 +100 200 ×1 +100 × 0.5 = 0.83 M. 200 +100 Balance the following equations : (a) H2O2 + MnO4–  Mn+2 + O2 (acidic medium) (b) Zn + HNO3 (dil)  Zn(NO3)2 + H2O + NH4NO3 (c) Cr3 + KOH + Cl2  K2CrO4 + KO4 + KCl + H2O. (d) P2H4  PH3 + P4 (e) Ca3 (PO4)2 + SiO2 + C  CaSiO3 + P4 + CO (a) 6H+ + 5H2O2 + 2MnO4–  2Mn+2 + 5O2 + 8H2O (b) 4Zn + 10HNO3 (dil)  4Zn(NO3)2 + 3H2O + NH4NO3 (c) 2Cr3 + 64KOH + 27Cl2  2K2CrO4 + 6KO4 + 54KCl + 32H2O. (d) 6P2H4  8PH3 + P4 (e) 2Ca3 (PO4)2 + 6SiO2 + 10C  6CaSiO3 + P4 + 10CO Ex. 20 Find the average and individual oxidation number of Fe & Pb in Fe3O4 & Pb3O4, which are mixed oxides. Sol. (i) Fe3O4 is mixture of FeO & Fe2O3 in 1 : 1 ratio so, individual oxidation number of Fe = +2 & +3 & average oxidation number = 1(+2) + 2(+3) = 8/3 3 (ii) Pb3O4 is a mixture of PbO & PbO2 in 2 : 1 molar ratio so, individual oxidation number of Pb are +2 & +4 & average oxidation number of Pb = Ex. 21 2(+2) +1(+4) =8/3 3 Calculate individual oxidation number of each S-atom in Na2S4O6 (sodium tetrathionate) with the help of its structure Sol. Ex. 22 518 gm of an aqueous solution contains 18 gm of glucose (mol.wt. = 180). What is the molality of the solution.  wt. of solvent = 518 – 18 = 500 gm. Ex. 23 0.25 of a substance is dissolved in 6.25 g of a solvent. Calculate the percentage amount of the substance in the solution. Sol. wt. of solution = 0.25 + 6.25 = 6.50. so % (w/w) = Ex. 24 A fresh H2O2 solution is labelled 11.2 V. This solution has the same concentration as a solution which is : (B) 3.4% (v / v) (C) 3.4% (w / v) Molarity of H2O2 = vol.strength 11.2 =1 = 11.2 11.2 Now, wt. of solute in g × 100 wt. of solution in mL %(w/v) = = Molarity × Mol. wt. of solute × = 1 × 34 × Ex. 25 molarity = 0.25 × 100 = 3.8%. 6.50 (A) 3.4% (w / w) Sol. So 18 / 180 = 0.2. 500 / 1000 Sol. 1 = 3.4% 10 (D) None of these 1 10 Ans. (C) A fresh H2O2 solution is labelled 11.2 V. This solution has the same concentration as a solution which is : (A) 3.4% (w / w) (B) 3.4% (v / v) (C) 3.4% (w / v) (D) None of these MOLE CONCEPT Sol. Molarity of H2O2 = vol.strength 11.2 = =1 11.2 11.2 Now, wt. of solute in g × 100 wt. of solution in mL %(w/v) = = Molarity × Mol. wt. of solute × = 1 × 34 × Ex. 26 Sol. 1 = 3.4% 10 1 10 Ans. (C) 0.25 of a substance is dissolved in 6.25 g of a solvent. Calculate the percentage amount of the substance in the solution. wt. of solution = 0.25 + 6.25 = 6.50. so % (w/w) = 0.25 × 100 = 3.8%. 6.50 Ex. 27 518 gm of an aqueous solution contains 18 gm of glucose (mol.wt. = 180). What is the molality of the solution. Sol. wt. of solvent = 518 – 18 = 500 gm.  Ex. 28 Ans. Calculate the resultant molarity of following : (a) 200 ml 1M HCl + 300 ml water (c) 200 ml 1M HCl + 100 ml 0.5 M H2SO4 (a) 0.4 M (b) 1.33 M (c) 1 M Sol. (a) Final molarity = so molarity = (b) 1500 ml 1M HCl + 18.25 g HCl (d) 200 ml 1M HCl + 100 ml 0.5 M HCl (d) 0.83 M. 200×1 + 0 = 0.4 M. 200 + 300 18.25 ×1000 36.5 = 1.33 M 1500 1500 ×1 + (b) Final molarity = (c) Final molarity of H+ = (d) Final molarity = 200 ×1 +100 × 0.5× 2 = 1 M. 200 +100 200 ×1 +100 × 0.5 = 0.83 M. 200 +100 18 / 180 = 0.2. 500 / 1000