Title 1.Basic Concepts of Chemistry (N.R.).pdf ✅

26 May 2009_____________________ Q.9 A solution contain 8 g of carbohydrate in 100 g of water has density 1.025 g/m and an osmotic pressure of 5 atm at 27oC the molar mass of carbohydrate is .......... (in g/mol). [0374]  Q.10 Calculate no. of mole in following – (a) 5.6 g CO2 (b) 74 g Ca (OH)2 (c) 16 g CH3OH (d) 30 g CH3COOH Q.11 Calculate no. of oxygen atom in following - (a) (NH4)2 Cr2O7 (2 mole) (b) H3PO4 (6.023 × 103 molecule) (c) H2S2O8(6.023 × 1022 molecule) (d) COOH | COOH (45 g) Q.12 Calculate total no. of atoms in following - (a) 0.3 mole (NH4)2Cr2O7 (b) 0.1 mole H2S2O8 (c) 0.9 mole H2SO5 (d) 90 g C6H12O6 18 Jun 2009 __________________________________ Q.13 If the components of air are N2, 78%; O2, 21% ; Ar, 0.9% and CO2, 0.1% by volume, what would be the molecular weight of air ? [28.964] Q.14 Oxygen is present in a 1-litre flask at a pressure of 7.6 × 10–10 mm of Hg at 0oC. Calculate the number of oxygen molecules in the flask. [2.65×1010] Q.15 A polystyrene, having the formula Br3C6H3(C3H8)n, was prepared by heating styrene with tribromobenzoyl peroxide in the absence of air. If it was found to contain 10.46% bromine by weight, find the value of n. [45] Q.16 0.75 mole of solid 'A4' and 2 moles of gaseous O2 are heated in a sealed vessel, completely using up the reactants and producing only one compound. It is found that when the temperature is reduced to the initial temperature, the contents of the vessel exhibit a pressure equal to half the original pressure. What conclusions can be drawn from these data about the product of the reaction ? [empirical formula of product is A3O4] Q.17 A compound which contains one atom of X and two atoms of Y for each three atoms of Z is made by mixing 5.0 g of X, 1.15 × 1023 atoms of Y and 0.03 mole of Z atoms. Given that only 4.40 g of the compound is formed. Calculate the atomic weight of Y if the atomic weights of X and Z are 60 and 80 amu respectively. [70 amu] Q.18 Two elements A (at. wt. = 75) and B(at. wt. = 16) combine to give a compound having 75.8% of A. What is the formula of the compound ? [A2B3] Q.19 116 g of Fe3O4 has 1.5 moles of Fe. Calculate the molecular weight of Fe3O4 without using atomic weights of Fe and O. [232] Q.20 In a Victor Meyer apparatus 0.168 g of a volatile compound displaced 49.4 mL of air measured over water at 20oC and 740 mm of pressure. Calculate the molecular weight of the compound. (Aqueous tension at 20oC = 18 mm) [86.06] Q.21 When 2 g of a gas 'A' is introduced into an evacuated flask kept at 25oC, the pressure is found to be one atmosphere. If 3 g of another gas 'B' is then added to the same flask, the total pressure becomes 1.5 atm. Assuming ideal gas behaviour, calculate the ratio of the molecular weights, MA : MB. [1/3] 19 Jun 09 _____________________________________ Q.22 What amount of CaO will be produced by 1 g of calcium ? [1.4 g] Q.23 A sample of KClO3 on decomposition yielded 448 mL of oxygen gas at NTP. Calculate (i) weight of oxygen produced, (ii) weight of KClO3 originally taken, and (iii) weight of KCl produced. (K = 39, Cl = 35.5 and O = 16) [ (i) 0.64 g (ii) 1.634 g (iii) 0.9937 g] Q.24 27.6 g of K2CO3 was treated by a series of reagents so as to convert all of its carbon to K2Zn3 [Fe(CN)6]2. Calculate the weight of the product.
Ans. [11.6 g] Q.25 1.0 g of an alloy of Al and Mg when treated with excess of dil. HCl forms MgCl2, AlCl3 and hydrogen. The evolved hydrogen, collected over Hg at 0oC, has a volume of 1.20 litres at 0.92 atm pressure. Calculate the composition of the alloy. (Al = 27 and Mg = 24) [% of Al = 55 & % of Mg = 45] Q.26 A sample of mixture of CaCl2 and NaCl weighing 4.22 g was treated to precipitate all the Ca as CaCO3, which was then heated and quantitatively converted to 0.959 g of CaO. Calculate the percentage of CaCl2 in the mixture. (Ca = 40, O = 16, C = 12 and Cl = 35.5) [45.04%] Q.27 A 2-g sample containing Na2CO3 and NaHCO3 loses 0.248 g when heated to 300oC, the temperature at which NaHCO3 decomposes to Na2CO3, CO2 and H2O. What is the percentage of Na2CO3 in the given mixture ? (Na = 23, C = 12, O = 16 and H =1) [66.4%] Q.28 For the production of equal amounts of hydrogen from the following reactions, which metal, Zn or Al, is less expensive if Zn costs about half as much as Al on a mass basis and by how much ? Zn + 2HCl → ZnCl2 + H2 2Al + 6HCl → 2AlCl3 + 3H2 [Al is less expensive by 44.61 %] Q.29 From the following reaction sequence, CaC2 + H2O → CaO + C2H2 C2H2 + H2 → C2H4 n C2H4 → (C2H4)n Calculate the mass of polyethylene which can be produced from 10 kg of CaC2. [4375 g] Q.30 From the following series of reactions, Cl2 + 2KOH → KCl + KClO + H2O 3KClO → 2KCl + KClO3 4KClO3 → 3KClO4 + KCl calculate the mass of chlorine needed to produce 100 g of KClO4. [204.5 g] 3 July 09______________________________________ Q.31 For the hydrolysis of 1 mole of CaH2 the number of moles of H2O required ........... [2] Sol. CaH2 + 2H2O → Ca(OH)2 + H2 10 July 09 ___________________________________ Q.32 A certain compound was known to have a formula which could be represented as [Pd (Cx HyNz)](ClO4)2 and analysis showed that the compound contain 30.15 % carbon and 5 % hydrogen. When the above compound is converted to the corresponding thiocyanate of molecular formula [Pd ( Cx Hy Nz)] (SCN)2 the analysis showed it contain 40.46 % carbon and 5.94 % hydrogen. Determine the value of x, y and z in nearest possible integers. At wt of Pd = 106 [x = 14 ; y = 28 ; z = 4] Sol. Let mol.wt of [ Pd (Cx HyNz)](ClO4)2 = M  M 12 x × 100 = 30.15 -- -(I) & M y × 100 = 5 ------(II) Mol wt of [Pd (Cx HyNz)](SCN)2 = M – 199 + 116 = M – 83  (M 83) 12 x 24 − + × 100 = 40.46 ---(III) and (M 83) y 100 −  = 5.94 --- (IV)  M 83 12 x 24 − + = 0.4046 or 83 30.15 1200 x 12 x 24 − + = 0.4046 or x = 14 M = 30.15 1200 x = 557 and y = 28 106 + 12 × 14 + 1 × 28 + 14 × z + 199 = 557 or 501 + 14 z = 557 or z = 4 x = 14 ; y = 28 & z = 4 Q.33 The ratio of concentration of NH3 in H2O to the concentration of NH3 in CHCl3 at 290 K is 24 at any pressure. Through one litre of pure water ammonia gas was passed , the concentration of
NH3 was measured to be 0.38 (M). To this solution 0.025 moles of CuSO4 was added. In NH3 solution all Cu2+ ions form the complex Cu 2+ 3 x (NH ) . Now over this solution one litre of CHCl3 was added , thoroughly shaken and equilibrated. In CHCl3 layer, Cu 2+ 3 x (NH ) was absent but in CHCl3 the concentration of NH3 was reduced to 0.0112 (M) at 290 K. Determine the value of x in nearest possible integers. [4] Sol. Initial moles of NH3 = 0.38 moles Moles of NH3 remaining after complexation = (0.38 – 0.025x) 3 3 3 Moles of NH in CHCl Moles of NH in water = 24  Moles of NH3 in CHCl3 = 24 Moles of NH in water 3 = 0.0112  0.38 – 0.025 x = 0.0112 × 24 + 0.0112 = 0.0112 × 25  0.025 x = 0.38 – 0.0112 × 25 = 0.38 – 0.28 = 0.1 or x = 0.025 0.1 = 4 Q.34 The vapour density of mixture containing NO2 (g) and N2O4 (g) is 38.3 at 27oC. Determine the number of millimoles of NO2 (g) present in 100 gm of the given mixture in nearest possible integers. [437] Sol. Let in 100 gm of mix , the NO2 (g) = x gm  N2O4 (g) present = (100 – x) gm.  total no. of moles present = 46 x + 92 (100 − x)  avg. mol wt of mix = 92 (100 x) 46 x 100 − + = 38.3 × 2  46 x + 92 (100 − x) = 2 38.3 100  = 38.3 50 or 92 (100 + x) = 38.3 50 or 100 + x = 38.3 50  92 = 120.1 or x = 20.1  no. of moles of NO2 present = 46 20.1 = 0.437 = 437 millimoles Q.35 A polystyrene having molecular formula Br3 (C6H3) (C8H8)n was prepared by heating styrene with tribromobenzoyl peroxide in the absence of air. If it was found to contain 10.46% bromine (At wt = 80) by mass, determine the value of ‘n’ in nearest possible integers. [19] Sol. Mol wt of the polymer = 3 × 80 + 75 + 104 n = 315 + 104 n  (315 104n) 240 + × 100 = 10.46 or 315 + 104 n = 0.1046 240 = 2294.5 or n = 19 Q.36 A solution of fatty acid “X” (M = 252 gm/mol) in benzene contain 4.2 gm of acid per litre. When this solution is dropped in a water surface the benzene evaporates and fatty acid “X” forms a monolayer film.of solid type. If we wish to cover 10 m2 area with monolayer film of fatty acid “X” , what volume of solution in ml should we use ? Each fatty acid molecule occupy 0.2 nm2 area. NA = 6 × 1023 [5] Sol. The no. of fatty acid molecule required = 18 0.2 10 10 −  = 5 × 1019  The no. of moles of fatty acid “X” required = 6 5 × 10–4