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68NUMIIIICAII'HVSKALCLLLMISTRY >- Problem 71. Find the milli-cquivalcnt ol : (a) Ca(OH) 2  in 111 g, (b) NaOH in 30  g,  (c) 11 2 S0 4  in 4.9 g. >ã Problem 72.  Find the weight of NaOH in its 60 milli-equivalcnts. >ã Problem 73.  Find the normality of H 2 S0 4  having 50 milli-equivalents in 3 litre. >- Problem 74.  Find the weight of H 2 S0 4  in 1200 mL of a solution of 0.4  N   strength. Problem 75.  Calculate normality of mixture obtained by mixing : (a) 100 mL of   0.1  NH 2 S0 4  + 50 mL of 0.25  N   NaOH. (b) 100 mL of 0.2 MH 2 S0 4  + 200 mL of 0.2 M HCI. (c) 100 mL of 0.2 A/H 2 S0 4  + 100 mL of 0.2 M NaOH. (d) 1 g equivalent of NaOH + 100 mL of   0.1  N   HCI. >ã Problem 76.  What volume of water is required to make 0.20  N   solution from 1600 mL of 0.2050  N   solution? >ã Problem 77.  How many mL of 2.0 AfPb(N0 3 ) 2  contains 600 mg Pb 2+ . >ã Problem 78.  How would you prepare exactly 3.0 litre of 1.0 MNaOH by mixing  proportions of stock solutions of 2.50 MNaOH and 0.40 M NaOH.  No water is to be used. > Problem 79.  What weight of Na 2 C0 3  of 95% purity would be required to neutralize 45.6 mL of 0.235  N   acid? > Problem 80.  Calculate normality of NH 4 OH when 2 g is present in 800 mL solution. Also calculate its molarity. >ã Problem 81.  What is the strength in g per litre of a solution of H 2 S0 4 , 12 ml ol which neutralized 15 mL of M10 NaOH solution? >> Problem 82.  Calculate the concentration of a solution obtained by mixing 300  g 25% by weight solution of NH 4 CI and 150 g of 40% by weight solution of NH 4 C1. >> Problem 83.  A sample of NaOH weighing 0.38 g is dissolved in water and the solution is made to 50.0 mL in a volumetric flask. What is the molarity of the resulting solution? >ã Problem 84.  How many moles of NaOH are contained in 27 mL of 0.15 MNaOH? >> Problem 85.  A sample of NaN0 3  weighing 0.38 g is placed in a 50.0 mL volumetric flask. The flask is then filled with water to the mark on the neck. What is the molarity of the solution? Problem  86. In a reaction vessel 0.184 g of NaOH is required to be added for completing the reaction. How many millilitre of 0.150 A/NaOII solution should be added for this requirement. >ã Problem 87.  Commercially available concentrated hydrochloric acid contains 38% HCI by mass. (a) What is the molarity of this solution? The density is 1.19 g mL -1 . (b) What volume of concentrated HCI is required to make 1.00 litre of 0.10 MHCI?  MOLEANDEQUIVALENTCONCIPT69 >ã Problem  88. Concentrated nitric acid used in the laboratory work is 68% nitric acid by mass in aqueous solution. What should be the molarity of such a sample of the acid if the density of solution is 1.504 g ml , 1 ? >- Problem  89. A solution of glucose in water is labelled as 10 per cent w/w, what would be the molality and mole fraction of each component in the £ solution? If the density of the solution is 1.2 g mL -1 , then what shall  be the molarity of the solution? >ã Problem  90. An antifreeze solution is prepared from 222.6 g of ethylene glycol [C 2 H 4 (OH) 2 ] and 200 g of water. Calculate the molality of the solution. If the density of the solution is 1.072 g mL -1  then what shall be the molarity of the solution? ã Problem 91. Calculate the amount of oxalic acid (H 2 C 2 0 4 .2H 2 0) required to obtain 250 mL of deci-molar solution. ã Problem 92. 4 g of NaOH are present in 0.1 dm 3  solution have specific gravity 1.038 g/mL. Calculate : (a) mole fraction of NaOH; (b) molality of NaOH solution; (c) molarity of NaOH solution; (d) normality of NaOH solution. ã Problem 93. Suppose 5 g of acetic acid are dissolved in one litre of ethanol. Assume < no reaction in between them. Calculate molality of resulting solution if density of ethanol is 0.789 g/mL. >> Problem 94. Find the molality of H 2 S0 4  solution whose specific gravity is 1.98 g mL 1  and 90% by volume H 2 S0 4 . ã Problem 95. A sample of H 2 S0 4  (density 1.787 g mL -1 ) is labelled as 80% by weight. What is molarity of acid? What volume of acid has to be used to make 1 litre of 0.2 MH 2 S0 4 ? ã Problem 96. 30 mL of   0.2  NBaCl 2  is mixed with 40 mL of   0.3  N A1 2 (S0 4 ) 3 . How many g of BaS0 4  are formed? >- Problem 97. 20 mL of 0.2 MAI 2 (S0 4 ) 3  mixed with 20 mL of 0.6 MBaCl 2 . Calculate the concentration of each ion in solution. >ã Problem 98. What are the final concentrations of all the ions when following are mixed? 50 mL of 0.12 A/ Fe(N0 3 ) 3 , 100 mL of 0.10A/FeCl 3  and 100 mL of 0.26M   Mg(N0 3 ) 2  . >ã Problem 99. 30 mL of 0.1 MBaCl 2  is mixed with 40 mL of 0.2 MAI 2 (S0 4  j 3 . What is the weight of BaS0 4  formed? BaCl 2  + Al 2 (S0 4 ) 3  > BaS0 4  + A1C1 3 ^Problem 100.  Calcium carbonate reacts with aqueous HCI to give CaCl 2  and C0 2 according to the reaction; CaC0 3 (s) + 2HCI(aq.) ã CaCl 2 (aq.) + C0 2 (g) + H 2 Q(I) What mass of CaC0 3  is required to react completely with 25 mL ol 0.75 A/HCI?  70'NUMERICALPHYSICALCHEMISTRY >*Problem 101.  Calculate the volume of 1.00 mol L  1  aqueous sodium hydroxide that is neutralized by 200 mL of 2.00 mol L -1  aqueous hydrochloric acid and the mass of sodium chloride produced. Neutralization reaction is;  NaOH(aq.) + HCl(aq.) > NaCI(aq.) + H 2 0(1). >>Problem 102.  How many mL of a 0. 1  M   HCI are required to react completely with 1 g mixture of Na 2 C0 3  and NaHC0 3  containing equimolar amounts of two? >-Problem 103.  A sample of drinking water was found to be severely contaminated with chloroform, CHC1 3 , supposed to be carcinogen. The level of contamination was 15 ppm (by mass). (i) Express this in per cent by mass. (ii) Determine the molality of chloroform in the water sample. >-Problem 104.  An aqueous solution of sodium chloride is marked 10% (w/w) on the  bottle. The density of the solution is 1.071 gml. -1 . What is its molality and molarity? Also, what is the mole fraction of each component in the solution?  72NUMERICALPHYSICALCHEMISTRY 61. 151.20 kg; 62. 5.18 g, Fe 2 0 3 ; 63. FeO = 21.06%, Fe 3 0 4  = 78.94%; 64. Al = 1.25 g, Zn = 0.42 g; 65. 8.39 g; 66. 1.345 x 10 21  atoms of Mg; 67. CaCl 2 .6H 2 0 = 9.9 g, H 2 0 = 90.1 | g;  68. 33.57%; 69. 27.52 ; 70. C 3 H 8 71. (a) 3000, (b) 750, (c) 100; 72. 2.4 g; 73. 0.0166; 74. 23.52 g; 75. (a) 0.0167, (b) 0.267, (c)0.1, (d) 9.9; 76. 40 mL; 77. 1.44 mL; 78. 857.14 mL of 2.50 M* 2142.96 mL of 0.4 M; 79. 0.5968 g; 80.  N   =  0.07,  M  =0.07; 81. 6.125 g/litre; 82. 30%; 83. 0.19; 84. 4.05 x 10~ 3 ; 85. 8.94 x 10~ 2 ; 86. 30.67 mL; 87. (a) 12.4, (b) 8.06 mL; 88. 16.23  M; 89. 0.617  m,  0.67 M, 0.011,0.989; 90. 17.95  m,  9.11  M\ 91. 3.15 g; 92. (a) 0.018, (b) 1.002  m  (c) 1  M,  (d)  1  N; 93. 0.1056; 94. 8.50; 95. 14.59, 13.71 mL; 96. 0.70 g; 97. Al 3+  = 0.2 M, CI = 0.6 M; 98. [Fe 3+ ] = 0.064M; [NOf] = 0.28M; [CP] = 0.12  M\ [Mg 2+ ] = 0.104M 99. 0.699 g; 100. 0.94 g; 101. 400 mL, 23.4 g; 102. 157.8 mL; 103. (i) 1.5 x 10 3 , (ii) 1.25 x 10^  m 104. M= 1.83,  m  = 1.90, m.f.  Naa  = 0.03, m.f. H I  2  O  = 0-97

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