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CLASS : XIth SUBJECT : CHEMISTRY DATE : DPP No. : 7 1 (d) Charcoal adsorbs gases. 2 (c) Given, V1 = 500 mL, T1 = 27 + 273 = 300 K V2 = ?, T2 = 42 + 273 = 315 K From Charles’ law V1T2 = V2T1 ∴ V2 = 500 × 315 300 = 525 mL Hence, increase in volume = 525 ― 500 = 25 mL 4 (d) CO reacts with red colouring haemoglobin molecules in blood to form a complex of cherry red colour. 5 (b) AgBr exhibits Frenkel defect due to large difference in the size of Ag + and Br ― ions 7 (b) The internal energy, i.e., kinetic energy of gas depends only on temperature. 8 (b) urms = 3RT M 9 (c) SO2 has higher value of van der Waals’ forces of attraction and thus, more easily liquefied. 10 (c) Liquefaction of a gas is easier if it possesses high Tc and higher Ti 11 (c) Topic :- STATES OF MATTER Solutions
1 V (mol litre ―1 ) Van der Waals’ equation for 1 mol of real gas is [P + a V 2][V ― b] = RT Given that b = 0 ∴ (P + a V 2)(V) = RT ∴ PV = RT ― a V ....(i) Following y = mx + c for the curve PV vs 1 V Slope = ―a Slope = 21.6 ― 20.1 2 ― 3 = ―1.5 ∴ a = 1.5 13 (b) Gases for which intermolecular forces of attraction are small such as N2, O2 etc have low value of Tc, thus liquefied above critical temperature 14 (b) d1T1 = d2T2 When p remains constant d1 = 16;d2 = 14;T1 = 273 K, T2 = ? d1T1 = d2T2 16 × 273 = 14 × T2 T2 = 312 K T2 = 312 ― 273 = 39°C 15 (c) d = Pm RT 0 2.0 3.0 24.6 23.1 21.6 20.1 P V (litre-atm m ol -1 [Graph not to scale]
16 (a) Both CO2 and N2O have same mol. wt. 17 (d) Mole fraction of nitrogen in air is greater than the given gases so it has highest partial pressure in the atmosphere. 18 (b) In rock salt structure, the coordination number of Na +:CI ― is 6 : 6 19 (c) CO2 is more easily liquefied than O2 gas. Hence (a) of CO2 is more than that of O2. Also CH4 is easily liquefied than H2 and He. Hence ′a′ of CH4 is more than H2 and He. He H2 O2 CO2 CH4 a 0.434 0.244 1.36 3.59 2.25 atom l 2 mole―2 b 0.0237 0.0266 0.0318 0.0427 0.0428 l mol―1 ∴ Order of a CH4 > O2 > H2 ∴ Order of b He < H2 < O2 < CO2 20 (a) (Tf ) irrev > (Tf ) rev

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